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2
problems/0001.两数之和.md
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@ -41,7 +41,7 @@
那么我们就应该想到使用哈希法了。
因为本地，我们不仅要知道元素有没有遍历过，还有知道这个元素对应的下标，**需要使用 key value结构来存放，key来存元素，value来存下标，那么使用map正合适**。
因为本地，我们不仅要知道元素有没有遍历过，还要知道这个元素对应的下标，**需要使用 key value结构来存放，key来存元素，value来存下标，那么使用map正合适**。
再来看一下使用数组和set来做哈希法的局限。

4
problems/0035.搜索插入位置.md
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@ -191,8 +191,8 @@ public:
};
```
* 时间复杂度：$O(\log n)$
* 时间复杂度：$O(1)$
* 时间复杂度：O(log n)
* 空间复杂度：O(1)
## 总结

146
problems/0098.验证二叉搜索树.md
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@ -371,117 +371,113 @@ class Solution {
## Python
**递归** - 利用BST中序遍历特性,把树"压缩"成数组
递归法（版本一）利用中序递增性质，转换成数组
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isValidBST(self, root: TreeNode) -> bool:
# 思路: 利用BST中序遍历的特性.
# 中序遍历输出的二叉搜索树节点的数值是有序序列
candidate_list = []
def __init__(self):
self.vec = []
def __traverse(root: TreeNode) -> None:
nonlocal candidate_list
if not root:
def traversal(self, root):
if root is None:
return
__traverse(root.left)
candidate_list.append(root.val)
__traverse(root.right)
self.traversal(root.left)
self.vec.append(root.val) # 将二叉搜索树转换为有序数组
self.traversal(root.right)
def __is_sorted(nums: list) -> bool:
for i in range(1, len(nums)):
if nums[i] <= nums[i - 1]: # ⚠️ 注意: Leetcode定义二叉搜索树中不能有重复元素
def isValidBST(self, root):
self.vec = [] # 清空数组
self.traversal(root)
for i in range(1, len(self.vec)):
# 注意要小于等于，搜索树里不能有相同元素
if self.vec[i] <= self.vec[i - 1]:
return False
return True
__traverse(root)
res = __is_sorted(candidate_list)
return res
```
**递归** - 标准做法
递归法（版本二）设定极小值，进行比较
```python
class Solution:
def isValidBST(self, root: TreeNode) -> bool:
# 规律: BST的中序遍历节点数值是从小到大.
cur_max = -float("INF")
def __isValidBST(root: TreeNode) -> bool:
nonlocal cur_max
def __init__(self):
self.maxVal = float('-inf') # 因为后台测试数据中有int最小值
if not root:
def isValidBST(self, root):
if root is None:
return True
is_left_valid = __isValidBST(root.left)
if cur_max < root.val:
cur_max = root.val
left = self.isValidBST(root.left)
# 中序遍历，验证遍历的元素是不是从小到大
if self.maxVal < root.val:
self.maxVal = root.val
else:
return False
is_right_valid = __isValidBST(root.right)
right = self.isValidBST(root.right)
return left and right
return is_left_valid and is_right_valid
return __isValidBST(root)
```
**递归** - 避免初始化最小值做法：
递归法（版本三）直接取该树的最小值
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isValidBST(self, root: TreeNode) -> bool:
# 规律: BST的中序遍历节点数值是从小到大.
pre = None
def __isValidBST(root: TreeNode) -> bool:
nonlocal pre
def __init__(self):
self.pre = None # 用来记录前一个节点
if not root:
def isValidBST(self, root):
if root is None:
return True
is_left_valid = __isValidBST(root.left)
if pre and pre.val>=root.val: return False
pre = root
is_right_valid = __isValidBST(root.right)
left = self.isValidBST(root.left)
if self.pre is not None and self.pre.val >= root.val:
return False
self.pre = root # 记录前一个节点
right = self.isValidBST(root.right)
return left and right
return is_left_valid and is_right_valid
return __isValidBST(root)
```
迭代法
```python
迭代-中序遍历
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isValidBST(self, root: TreeNode) -> bool:
def isValidBST(self, root):
stack = []
cur = root
pre = None
while cur or stack:
if cur: # 指针来访问节点，访问到最底层
pre = None # 记录前一个节点
while cur is not None or len(stack) > 0:
if cur is not None:
stack.append(cur)
cur = cur.left
else: # 逐一处理节点
cur = stack.pop()
if pre and cur.val <= pre.val: # 比较当前节点和前节点的值的大小
cur = cur.left # 左
else:
cur = stack.pop() # 中
if pre is not None and cur.val <= pre.val:
return False
pre = cur
cur = cur.right
pre = cur # 保存前一个访问的结点
cur = cur.right # 右
return True
```
```python
# 遵循Carl的写法，只添加了节点判断的部分
class Solution:
def isValidBST(self, root: TreeNode) -> bool:
# method 2
que, pre = [], None
while root or que:
while root:
que.append(root)
root = root.left
root = que.pop()
# 对第一个节点只做记录，对后面的节点进行比较
if pre is None:
pre = root.val
else:
if pre >= root.val: return False
pre = root.val
root = root.right
return True
```
## Go

38
problems/0106.从中序与后序遍历序列构造二叉树.md
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@ -400,8 +400,6 @@ public:
};
```
## Python
# 105.从前序与中序遍历序列构造二叉树
@ -692,38 +690,6 @@ class Solution {
## Python
```python
class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
# 第一步: 特殊情况讨论: 树为空. 或者说是递归终止条件
if not postorder:
return
# 第二步: 后序遍历的最后一个就是当前的中间节点
root_val = postorder[-1]
root = TreeNode(root_val)
# 第三步: 找切割点.
root_index = inorder.index(root_val)
# 第四步: 切割inorder数组. 得到inorder数组的左,右半边.
left_inorder = inorder[:root_index]
right_inorder = inorder[root_index + 1:]
# 第五步: 切割postorder数组. 得到postorder数组的左,右半边.
# ⭐️ 重点1: 中序数组大小一定跟后序数组大小是相同的.
left_postorder = postorder[:len(left_inorder)]
right_postorder = postorder[len(left_inorder): len(postorder) - 1]
# 第六步: 递归
root.left = self.buildTree(left_inorder, left_postorder)
root.right = self.buildTree(right_inorder, right_postorder)
# 第七步: 返回答案
return root
```
105.从前序与中序遍历序列构造二叉树
```python
@ -752,7 +718,7 @@ class Solution:
# 第六步: 递归
root.left = self.buildTree(preorder_left, inorder_left)
root.right = self.buildTree(preorder_right, inorder_right)
# 第七步: 返回答案
return root
```
@ -784,7 +750,7 @@ class Solution:
# 第六步: 递归
root.left = self.buildTree(inorder_left, postorder_left)
root.right = self.buildTree(inorder_right, postorder_right)
# 第七步: 返回答案
return root
```

90
problems/0108.将有序数组转换为二叉搜索树.md
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@ -316,73 +316,65 @@ class Solution {
```
## Python
**递归**
递归法
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
'''
构造二叉树：重点是选取数组最中间元素为分割点，左侧是递归左区间；右侧是递归右区间
必然是平衡树
左闭右闭区间
'''
# 返回根节点
root = self.traversal(nums, 0, len(nums)-1)
return root
def traversal(self, nums: List[int], left: int, right: int) -> TreeNode:
# Base Case
if left > right:
return None
# 确定左右界的中心，防越界
mid = left + (right - left) // 2
# 构建根节点
mid_root = TreeNode(nums[mid])
# 构建以左右界的中心为分割点的左右子树
mid_root.left = self.traversal(nums, left, mid-1)
mid_root.right = self.traversal(nums, mid+1, right)
root = TreeNode(nums[mid])
root.left = self.traversal(nums, left, mid - 1)
root.right = self.traversal(nums, mid + 1, right)
return root
def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
root = self.traversal(nums, 0, len(nums) - 1)
return root
# 返回由被传入的左右界定义的某子树的根节点
return mid_root
```
**迭代**（左闭右开）
迭代法
```python
from collections import deque
class Solution:
def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
if len(nums) == 0: return None
root = TreeNode() # 初始化
nodeSt = [root]
leftSt = [0]
rightSt = [len(nums)]
def sortedArrayToBST(self, nums: List[int]) -> TreeNode:
if len(nums) == 0:
return None
while nodeSt:
node = nodeSt.pop() # 处理根节点
left = leftSt.pop()
right = rightSt.pop()
root = TreeNode(0) # 初始根节点
nodeQue = deque() # 放遍历的节点
leftQue = deque() # 保存左区间下标
rightQue = deque() # 保存右区间下标
nodeQue.append(root) # 根节点入队列
leftQue.append(0) # 0为左区间下标初始位置
rightQue.append(len(nums) - 1) # len(nums) - 1为右区间下标初始位置
while nodeQue:
curNode = nodeQue.popleft()
left = leftQue.popleft()
right = rightQue.popleft()
mid = left + (right - left) // 2
node.val = nums[mid]
if left < mid: # 处理左区间
node.left = TreeNode()
nodeSt.append(node.left)
leftSt.append(left)
rightSt.append(mid)
curNode.val = nums[mid] # 将mid对应的元素给中间节点
if right > mid + 1: # 处理右区间
node.right = TreeNode()
nodeSt.append(node.right)
leftSt.append(mid + 1)
rightSt.append(right)
if left <= mid - 1: # 处理左区间
curNode.left = TreeNode(0)
nodeQue.append(curNode.left)
leftQue.append(left)
rightQue.append(mid - 1)
if right >= mid + 1: # 处理右区间
curNode.right = TreeNode(0)
nodeQue.append(curNode.right)
leftQue.append(mid + 1)
rightQue.append(right)
return root
```
## Go

62
problems/0110.平衡二叉树.md
View File

@ -536,16 +536,14 @@ class Solution:
```python
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
return self.height(root) != -1
def height(self, node: TreeNode) -> int:
def isBalanced(self, root: Optional[TreeNode]) -> bool:
return self.get_hight(root) != -1
def get_hight(self, node):
if not node:
return 0
left = self.height(node.left)
if left == -1:
return -1
right = self.height(node.right)
if right == -1 or abs(left - right) > 1:
left = self.get_hight(node.left)
right = self.get_hight(node.right)
if left == -1 or right == -1 or abs(left - right) > 1:
return -1
return max(left, right) + 1
```
@ -553,6 +551,52 @@ class Solution:
迭代法：
```python
class Solution:
def getDepth(self, cur):
st = []
if cur is not None:
st.append(cur)
depth = 0
result = 0
while st:
node = st[-1]
if node is not None:
st.pop()
st.append(node) # 中
st.append(None)
depth += 1
if node.right:
st.append(node.right) # 右
if node.left:
st.append(node.left) # 左
else:
node = st.pop()
st.pop()
depth -= 1
result = max(result, depth)
return result
def isBalanced(self, root):
st = []
if root is None:
return True
st.append(root)
while st:
node = st.pop() # 中
if abs(self.getDepth(node.left) - self.getDepth(node.right)) > 1:
return False
if node.right:
st.append(node.right) # 右（空节点不入栈）
if node.left:
st.append(node.left) # 左（空节点不入栈）
return True
```
迭代法精简版：
```python
class Solution:
def isBalanced(self, root: Optional[TreeNode]) -> bool:
@ -576,8 +620,6 @@ class Solution:
height_map[real_node] = 1 + max(left, right)
return True
```
### Go
```Go

106
problems/0111.二叉树的最小深度.md
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@ -302,36 +302,72 @@ class Solution {
## Python
递归法：
递归法（版本一）
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def minDepth(self, root: TreeNode) -> int:
if not root:
def getDepth(self, node):
if node is None:
return 0
leftDepth = self.getDepth(node.left) # 左
rightDepth = self.getDepth(node.right) # 右
if not root.left and not root.right:
return 1
# 当一个左子树为空，右不为空，这时并不是最低点
if node.left is None and node.right is not None:
return 1 + rightDepth
left_depth = float('inf')
right_depth = float('inf')
# 当一个右子树为空，左不为空，这时并不是最低点
if node.left is not None and node.right is None:
return 1 + leftDepth
if root.left:
left_depth = self.minDepth(root.left)
if root.right:
right_depth = self.minDepth(root.right)
result = 1 + min(leftDepth, rightDepth)
return result
return 1 + min(left_depth, right_depth)
def minDepth(self, root):
return self.getDepth(root)
```
递归法（版本二）
迭代法：
```python
class Solution:
def minDepth(self, root):
if root is None:
return 0
if root.left is None and root.right is not None:
return 1 + self.minDepth(root.right)
if root.left is not None and root.right is None:
return 1 + self.minDepth(root.left)
return 1 + min(self.minDepth(root.left), self.minDepth(root.right))
```
递归法（版本三）前序
```python
class Solution:
def __init__(self):
self.result = float('inf')
def getDepth(self, node, depth):
if node is None:
return
if node.left is None and node.right is None:
self.result = min(self.result, depth)
if node.left:
self.getDepth(node.left, depth + 1)
if node.right:
self.getDepth(node.right, depth + 1)
def minDepth(self, root):
if root is None:
return 0
self.getDepth(root, 1)
return self.result
```
迭代法
```python
# Definition for a binary tree node.
@ -364,39 +400,7 @@ class Solution:
return depth
```
迭代法：
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def minDepth(self, root: TreeNode) -> int:
if not root:
return 0
queue = collections.deque([(root, 1)])
while queue:
node, depth = queue.popleft()
# Check if the node is a leaf node
if not node.left and not node.right:
return depth
# Add left and right child to the queue
if node.left:
queue.append((node.left, depth+1))
if node.right:
queue.append((node.right, depth+1))
return 0
```
## Go

47
problems/0226.翻转二叉树.md
View File

@ -991,6 +991,53 @@ impl Solution {
}
```
### C#
```csharp
//递归
public class Solution {
public TreeNode InvertTree(TreeNode root) {
if (root == null) return root;
swap(root);
InvertTree(root.left);
InvertTree(root.right);
return root;
}
public void swap(TreeNode node) {
TreeNode temp = node.left;
node.left = node.right;
node.right = temp;
}
}
```
```csharp
//迭代
public class Solution {
public TreeNode InvertTree(TreeNode root) {
if (root == null) return null;
Stack<TreeNode> stack=new Stack<TreeNode>();
stack.Push(root);
while(stack.Count>0)
{
TreeNode node = stack.Pop();
swap(node);
if(node.right!=null) stack.Push(node.right);
if(node.left!=null) stack.Push(node.left);
}
return root;
}
public void swap(TreeNode node) {
TreeNode temp = node.left;
node.left = node.right;
node.right = temp;
}
}
```
<p align="center">
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>

49
problems/0235.二叉搜索树的最近公共祖先.md
View File

@ -275,34 +275,57 @@ class Solution {
## Python
递归法：
递归法（版本一）
```python
class Solution:
"""二叉搜索树的最近公共祖先 递归法"""
def traversal(self, cur, p, q):
if cur is None:
return cur
# 中
if cur.val > p.val and cur.val > q.val: # 左
left = self.traversal(cur.left, p, q)
if left is not None:
return left
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if root.val > p.val and root.val > q.val:
return self.lowestCommonAncestor(root.left, p, q)
if root.val < p.val and root.val < q.val:
return self.lowestCommonAncestor(root.right, p, q)
return root
if cur.val < p.val and cur.val < q.val: # 右
right = self.traversal(cur.right, p, q)
if right is not None:
return right
return cur
def lowestCommonAncestor(self, root, p, q):
return self.traversal(root, p, q)
```
迭代法：
迭代法（版本二）精简
```python
class Solution:
"""二叉搜索树的最近公共祖先 迭代法"""
def lowestCommonAncestor(self, root, p, q):
if root.val > p.val and root.val > q.val:
return self.lowestCommonAncestor(root.left, p, q)
elif root.val < p.val and root.val < q.val:
return self.lowestCommonAncestor(root.right, p, q)
else:
return root
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
while True:
```
迭代法
```python
class Solution:
def lowestCommonAncestor(self, root, p, q):
while root:
if root.val > p.val and root.val > q.val:
root = root.left
elif root.val < p.val and root.val < q.val:
root = root.right
else:
return root
```
return None
```
## Go
递归法：

39
problems/0236.二叉树的最近公共祖先.md
View File

@ -274,25 +274,44 @@ class Solution {
```
## Python
递归法（版本一）
```python
class Solution:
"""二叉树的最近公共祖先 递归法"""
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if not root or root == p or root == q:
def lowestCommonAncestor(self, root, p, q):
if root == q or root == p or root is None:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right:
if left is not None and right is not None:
return root
if left:
return left
return right
```
if left is None and right is not None:
return right
elif left is not None and right is None:
return left
else:
return None
```
递归法（版本二）精简
```python
class Solution:
def lowestCommonAncestor(self, root, p, q):
if root == q or root == p or root is None:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left is not None and right is not None:
return root
if left is None:
return right
return left
```
## Go
```Go

114
problems/0257.二叉树的所有路径.md
View File

@ -470,70 +470,65 @@ class Solution {
## Python:
递归法+回溯（版本一）
递归法+回溯
```Python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
import copy
from typing import List, Optional
class Solution:
def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
if not root:
return []
result = []
self.generate_paths(root, [], result)
return result
def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None:
path.append(node.val)
if not node.left and not node.right:
result.append('->'.join(map(str, path)))
if node.left:
self.generate_paths(node.left, copy.copy(path), result)
if node.right:
self.generate_paths(node.right, copy.copy(path), result)
path.pop()
```
递归法+回溯（版本二）
```Python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
import copy
from typing import List, Optional
class Solution:
def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
if not root:
return []
result = []
self.generate_paths(root, [], result)
return result
def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None:
if not node:
def traversal(self, cur, path, result):
path.append(cur.val) # 中
if not cur.left and not cur.right: # 到达叶子节点
sPath = '->'.join(map(str, path))
result.append(sPath)
return
path.append(node.val)
if not node.left and not node.right:
if cur.left: # 左
self.traversal(cur.left, path, result)
path.pop() # 回溯
if cur.right: # 右
self.traversal(cur.right, path, result)
path.pop() # 回溯
def binaryTreePaths(self, root):
result = []
path = []
if not root:
return result
self.traversal(root, path, result)
return result
```
递归法+隐形回溯（版本一）
```Python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
from typing import List, Optional
class Solution:
def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
if not root:
return []
result = []
self.traversal(root, [], result)
return result
def traversal(self, cur: TreeNode, path: List[int], result: List[str]) -> None:
if not cur:
return
path.append(cur.val)
if not cur.left and not cur.right:
result.append('->'.join(map(str, path)))
else:
self.generate_paths(node.left, copy.copy(path), result)
self.generate_paths(node.right, copy.copy(path), result)
path.pop()
if cur.left:
self.traversal(cur.left, path[:], result)
if cur.right:
self.traversal(cur.right, path[:], result)
```
递归法+隐形回溯
递归法+隐形回溯（版本二）
```Python
# Definition for a binary tree node.
# class TreeNode:
@ -566,16 +561,11 @@ class Solution:
迭代法：
```Python
from collections import deque
class Solution:
"""二叉树的所有路径 迭代法"""
def binaryTreePaths(self, root: TreeNode) -> List[str]:
# 题目中节点数至少为1
stack, path_st, result = deque([root]), deque(), []
path_st.append(str(root.val))
stack, path_st, result = [root], [str(root.val)], []
while stack:
cur = stack.pop()

2
problems/0300.最长上升子序列.md
View File

@ -149,7 +149,7 @@ class Solution:
if len(nums) <= 1:
return len(nums)
dp = [1] * len(nums)
result = 0
result = 1
for i in range(1, len(nums)):
for j in range(0, i):
if nums[i] > nums[j]:

84
problems/0404.左叶子之和.md
View File

@ -247,8 +247,7 @@ class Solution {
### Python
**递归后序遍历**
递归
```python
# Definition for a binary tree node.
# class TreeNode:
@ -257,47 +256,64 @@ class Solution {
# self.left = left
# self.right = right
class Solution:
def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
if not root:
def sumOfLeftLeaves(self, root):
if root is None:
return 0
if root.left is None and root.right is None:
return 0
# 检查根节点的左子节点是否为叶节点
if root.left and not root.left.left and not root.left.right:
left_val = root.left.val
else:
left_val = self.sumOfLeftLeaves(root.left)
leftValue = self.sumOfLeftLeaves(root.left) # 左
if root.left and not root.left.left and not root.left.right: # 左子树是左叶子的情况
leftValue = root.left.val
# 递归地计算右子树左叶节点的和
right_val = self.sumOfLeftLeaves(root.right)
rightValue = self.sumOfLeftLeaves(root.right) # 右
return left_val + right_val
sum_val = leftValue + rightValue # 中
return sum_val
```
递归精简版
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def sumOfLeftLeaves(self, root):
if root is None:
return 0
leftValue = 0
if root.left is not None and root.left.left is None and root.left.right is None:
leftValue = root.left.val
return leftValue + self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)
```
**迭代**
迭代法
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def sumOfLeftLeaves(self, root: TreeNode) -> int:
"""
Idea: Each time check current node's left node.
If current node don't have one, skip it.
"""
stack = []
if root:
stack.append(root)
res = 0
def sumOfLeftLeaves(self, root):
if root is None:
return 0
st = [root]
result = 0
while st:
node = st.pop()
if node.left and node.left.left is None and node.left.right is None:
result += node.left.val
if node.right:
st.append(node.right)
if node.left:
st.append(node.left)
return result
while stack:
# 每次都把当前节点的左节点加进去.
cur_node = stack.pop()
if cur_node.left and not cur_node.left.left and not cur_node.left.right:
res += cur_node.left.val
if cur_node.left:
stack.append(cur_node.left)
if cur_node.right:
stack.append(cur_node.right)
return res
```
### Go

18
problems/0406.根据身高重建队列.md
View File

@ -295,19 +295,19 @@ var reconstructQueue = function(people) {
```Rust
impl Solution {
pub fn reconstruct_queue(people: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
let mut people = people;
pub fn reconstruct_queue(mut people: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
let mut queue = vec![];
people.sort_by(|a, b| {
if a[0] == b[0] { return a[1].cmp(&b[1]); }
if a[0] == b[0] {
return a[1].cmp(&b[1]);
}
b[0].cmp(&a[0])
});
let mut que: Vec<Vec<i32>> = Vec::new();
que.push(people[0].clone());
for i in 1..people.len() {
let position = people[i][1];
que.insert(position as usize, people[i].clone());
queue.push(people[0].clone());
for v in people.iter().skip(1) {
queue.insert(v[1] as usize, v.clone());
}
que
queue
}
}
```

168
problems/0450.删除二叉搜索树中的节点.md
View File

@ -324,88 +324,90 @@ class Solution {
```
## Python
递归法（版本一）
```python
class Solution:
def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
if not root : return None # 节点为空，返回
if root.val < key :
root.right = self.deleteNode(root.right, key)
elif root.val > key :
root.left = self.deleteNode(root.left, key)
def deleteNode(self, root, key):
if root is None:
return root
if root.val == key:
if root.left is None and root.right is None:
return None
elif root.left is None:
return root.right
elif root.right is None:
return root.left
else:
# 当前节点的左子树为空，返回当前的右子树
if not root.left : return root.right
# 当前节点的右子树为空，返回当前的左子树
if not root.right: return root.left
# 左右子树都不为空，找到右孩子的最左节点 记为p
node = root.right
while node.left :
node = node.left
# 将当前节点的左子树挂在p的左孩子上
node.left = root.left
# 当前节点的右子树替换掉当前节点，完成当前节点的删除
root = root.right
cur = root.right
while cur.left is not None:
cur = cur.left
cur.left = root.left
return root.right
if root.val > key:
root.left = self.deleteNode(root.left, key)
if root.val < key:
root.right = self.deleteNode(root.right, key)
return root
```
**普通二叉树的删除方式**
递归法（版本二）
```python
class Solution:
def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
if not root: return root
if root.val == key:
if not root.right: # 这里第二次操作目标值：最终删除的作用
return root.left
tmp = root.right
while tmp.left:
tmp = tmp.left
root.val, tmp.val = tmp.val, root.val # 这里第一次操作目标值：交换目标值其右子树最左面节点。
root.left = self.deleteNode(root.left, key)
root.right = self.deleteNode(root.right, key)
def deleteNode(self, root, key):
if root is None: # 如果根节点为空，直接返回
return root
if root.val == key: # 找到要删除的节点
if root.right is None: # 如果右子树为空，直接返回左子树作为新的根节点
return root.left
cur = root.right
while cur.left: # 找到右子树中的最左节点
cur = cur.left
root.val, cur.val = cur.val, root.val # 将要删除的节点值与最左节点值交换
root.left = self.deleteNode(root.left, key) # 在左子树中递归删除目标节点
root.right = self.deleteNode(root.right, key) # 在右子树中递归删除目标节点
return root
```
**迭代法**
```python
class Solution:
def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
# 找到节点后分两步，1. 把节点的左子树和右子树连起来，2. 把右子树跟父节点连起来
# root is None
if not root: return root
p = root
last = None
while p:
if p.val==key:
# 1. connect left to right
# right is not None -> left is None | left is not None
if p.right:
if p.left:
node = p.right
while node.left:
node = node.left
node.left = p.left
right = p.right
else:
# right is None -> right=left
right = p.left
# 2. connect right to last
if last==None:
root = right
elif last.val>key:
last.left = right
else:
last.right = right
# 3. return
def deleteOneNode(self, target: TreeNode) -> TreeNode:
"""
将目标节点（删除节点）的左子树放到目标节点的右子树的最左面节点的左孩子位置上
并返回目标节点右孩子为新的根节点
是动画里模拟的过程
"""
if target is None:
return target
if target.right is None:
return target.left
cur = target.right
while cur.left:
cur = cur.left
cur.left = target.left
return target.right
def deleteNode(self, root: TreeNode, key: int) -> TreeNode:
if root is None:
return root
cur = root
pre = None # 记录cur的父节点，用来删除cur
while cur:
if cur.val == key:
break
pre = cur
if cur.val > key:
cur = cur.left
else:
# Update last and continue
last = p
if p.val>key:
p = p.left
else:
p = p.right
cur = cur.right
if pre is None: # 如果搜索树只有头结点
return self.deleteOneNode(cur)
# pre 要知道是删左孩子还是右孩子
if pre.left and pre.left.val == key:
pre.left = self.deleteOneNode(cur)
if pre.right and pre.right.val == key:
pre.right = self.deleteOneNode(cur)
return root
```
@ -680,6 +682,40 @@ object Solution {
}
```
## rust
```rust
impl Solution {
pub fn delete_node(
root: Option<Rc<RefCell<TreeNode>>>,
key: i32,
) -> Option<Rc<RefCell<TreeNode>>> {
root.as_ref()?;
let mut node = root.as_ref().unwrap().borrow_mut();
match node.val.cmp(&key) {
std::cmp::Ordering::Less => node.right = Self::delete_node(node.right.clone(), key),
std::cmp::Ordering::Equal => match (node.left.clone(), node.right.clone()) {
(None, None) => return None,
(None, Some(r)) => return Some(r),
(Some(l), None) => return Some(l),
(Some(l), Some(r)) => {
let mut cur = Some(r.clone());
while let Some(n) = cur.clone().unwrap().borrow().left.clone() {
cur = Some(n);
}
cur.unwrap().borrow_mut().left = Some(l);
return Some(r);
}
},
std::cmp::Ordering::Greater => node.left = Self::delete_node(node.left.clone(), key),
}
drop(node);
root
}
}
```
<p align="center">
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>

142
problems/0501.二叉搜索树中的众数.md
View File

@ -528,8 +528,7 @@ class Solution {
## Python
> 递归法
> 常量空间，递归产生的栈不算
递归法（版本一）利用字典
```python
# Definition for a binary tree node.
@ -538,77 +537,108 @@ class Solution {
# self.val = val
# self.left = left
# self.right = right
from collections import defaultdict
class Solution:
def __init__(self):
self.pre = TreeNode()
self.count = 0
self.max_count = 0
self.result = []
def searchBST(self, cur, freq_map):
if cur is None:
return
freq_map[cur.val] += 1 # 统计元素频率
self.searchBST(cur.left, freq_map)
self.searchBST(cur.right, freq_map)
def findMode(self, root: TreeNode) -> List[int]:
if not root: return None
self.search_BST(root)
return self.result
def findMode(self, root):
freq_map = defaultdict(int) # key:元素，value:出现频率
result = []
if root is None:
return result
self.searchBST(root, freq_map)
max_freq = max(freq_map.values())
for key, freq in freq_map.items():
if freq == max_freq:
result.append(key)
return result
def search_BST(self, cur: TreeNode) -> None:
if not cur: return None
self.search_BST(cur.left)
# 第一个节点
if not self.pre:
self.count = 1
# 与前一个节点数值相同
elif self.pre.val == cur.val:
self.count += 1
# 与前一个节点数值不相同
else:
self.count = 1
self.pre = cur
if self.count == self.max_count:
self.result.append(cur.val)
if self.count > self.max_count:
self.max_count = self.count
self.result = [cur.val] # 清空self.result，确保result之前的的元素都失效
self.search_BST(cur.right)
```
递归法（版本二）利用二叉搜索树性质
> 迭代法-中序遍历
> 利用二叉搜索树特性，在历遍过程中更新结果，一次历遍
> 但需要使用额外空间存储历遍的节点
```python
class Solution:
def findMode(self, root: TreeNode) -> List[int]:
stack = []
def __init__(self):
self.maxCount = 0 # 最大频率
self.count = 0 # 统计频率
self.pre = None
self.result = []
def searchBST(self, cur):
if cur is None:
return
self.searchBST(cur.left) # 左
# 中
if self.pre is None: # 第一个节点
self.count = 1
elif self.pre.val == cur.val: # 与前一个节点数值相同
self.count += 1
else: # 与前一个节点数值不同
self.count = 1
self.pre = cur # 更新上一个节点
if self.count == self.maxCount: # 如果与最大值频率相同，放进result中
self.result.append(cur.val)
if self.count > self.maxCount: # 如果计数大于最大值频率
self.maxCount = self.count # 更新最大频率
self.result = [cur.val] # 很关键的一步，不要忘记清空result，之前result里的元素都失效了
self.searchBST(cur.right) # 右
return
def findMode(self, root):
self.count = 0
self.maxCount = 0
self.pre = None # 记录前一个节点
self.result = []
self.searchBST(root)
return self.result
```
迭代法
```python
class Solution:
def findMode(self, root):
st = []
cur = root
pre = None
maxCount, count = 0, 0
res = []
while cur or stack:
if cur: # 指针来访问节点，访问到最底层
stack.append(cur)
cur = cur.left
else: # 逐一处理节点
cur = stack.pop()
if pre == None: # 第一个节点
maxCount = 0 # 最大频率
count = 0 # 统计频率
result = []
while cur is not None or st:
if cur is not None: # 指针来访问节点，访问到最底层
st.append(cur) # 将访问的节点放进栈
cur = cur.left # 左
else:
cur = st.pop()
if pre is None: # 第一个节点
count = 1
elif pre.val == cur.val: # 与前一个节点数值相同
count += 1
else:
else: # 与前一个节点数值不同
count = 1
if count == maxCount:
res.append(cur.val)
if count > maxCount:
maxCount = count
res.clear()
res.append(cur.val)
if count == maxCount: # 如果和最大值相同，放进result中
result.append(cur.val)
if count > maxCount: # 如果计数大于最大值频率
maxCount = count # 更新最大频率
result = [cur.val] # 很关键的一步，不要忘记清空result，之前result里的元素都失效了
pre = cur
cur = cur.right
return res
cur = cur.right # 右
return result
```
## Go

15
problems/0513.找树左下角的值.md
View File

@ -330,19 +330,24 @@ class Solution:
# self.right = right
from collections import deque
class Solution:
def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
queue = deque([root])
def findBottomLeftValue(self, root):
if root is None:
return 0
queue = deque()
queue.append(root)
result = 0
while queue:
size = len(queue)
leftmost = queue[0].val
for i in range(size):
node = queue.popleft()
if i == 0:
result = node.val
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
if not queue:
return leftmost
return result
```

114
problems/0530.二叉搜索树的最小绝对差.md
View File

@ -237,66 +237,82 @@ class Solution {
```
## Python
递归
递归法（版本一）利用中序递增，结合数组
```python
class Solution:
def getMinimumDifference(self, root: TreeNode) -> int:
res = []
r = float("inf")
def buildaList(root): //把二叉搜索树转换成有序数组
if not root: return None
if root.left: buildaList(root.left) //左
res.append(root.val) //中
if root.right: buildaList(root.right) //右
return res
def __init__(self):
self.vec = []
buildaList(root)
for i in range(len(res)-1): // 统计有序数组的最小差值
r = min(abs(res[i]-res[i+1]),r)
return r
class Solution: # 双指针法，不用数组 (同Carl写法) - 更快
def getMinimumDifference(self, root: Optional[TreeNode]) -> int:
global pre,minval
pre = None
minval = 10**5
self.traversal(root)
return minval
def traversal(self,root):
global pre,minval
if not root: return None
def traversal(self, root):
if root is None:
return
self.traversal(root.left)
if pre and root.val-pre.val<minval:
minval = root.val-pre.val
pre = root
self.vec.append(root.val) # 将二叉搜索树转换为有序数组
self.traversal(root.right)
```
迭代法-中序遍历
```python
class Solution:
def getMinimumDifference(self, root: TreeNode) -> int:
stack = []
cur = root
pre = None
def getMinimumDifference(self, root):
self.vec = []
self.traversal(root)
if len(self.vec) < 2:
return 0
result = float('inf')
while cur or stack:
if cur: # 指针来访问节点，访问到最底层
stack.append(cur)
cur = cur.left
else: # 逐一处理节点
cur = stack.pop()
if pre: # 当前节点和前节点的值的差值
result = min(result, abs(cur.val - pre.val))
pre = cur
cur = cur.right
for i in range(1, len(self.vec)):
# 统计有序数组的最小差值
result = min(result, self.vec[i] - self.vec[i - 1])
return result
```
递归法（版本二）利用中序递增，找到该树最小值
```python
class Solution:
def __init__(self):
self.result = float('inf')
self.pre = None
def traversal(self, cur):
if cur is None:
return
self.traversal(cur.left) # 左
if self.pre is not None: # 中
self.result = min(self.result, cur.val - self.pre.val)
self.pre = cur # 记录前一个
self.traversal(cur.right) # 右
def getMinimumDifference(self, root):
self.traversal(root)
return self.result
```
迭代法
```python
class Solution:
def getMinimumDifference(self, root):
stack = []
cur = root
pre = None
result = float('inf')
while cur is not None or len(stack) > 0:
if cur is not None:
stack.append(cur) # 将访问的节点放进栈
cur = cur.left # 左
else:
cur = stack.pop()
if pre is not None: # 中
result = min(result, cur.val - pre.val)
pre = cur
cur = cur.right # 右
return result
```
## Go：
中序遍历，然后计算最小差值

51
problems/0538.把二叉搜索树转换为累加树.md
View File

@ -200,8 +200,30 @@ class Solution {
```
## Python
**递归**
递归法(版本一)
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def convertBST(self, root: TreeNode) -> TreeNode:
self.pre = 0 # 记录前一个节点的数值
self.traversal(root)
return root
def traversal(self, cur):
if cur is None:
return
self.traversal(cur.right)
cur.val += self.pre
self.pre = cur.val
self.traversal(cur.left)
```
递归法（版本二）
```python
# Definition for a binary tree node.
# class TreeNode:
@ -234,7 +256,32 @@ class Solution:
return root
```
**迭代**
迭代法（版本一）
```python
class Solution:
def __init__(self):
self.pre = 0 # 记录前一个节点的数值
def traversal(self, root):
stack = []
cur = root
while cur or stack:
if cur:
stack.append(cur)
cur = cur.right # 右
else:
cur = stack.pop() # 中
cur.val += self.pre
self.pre = cur.val
cur = cur.left # 左
def convertBST(self, root):
self.pre = 0
self.traversal(root)
return root
```
迭代法（版本二）
```python
class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:

67
problems/0617.合并二叉树.md
View File

@ -352,8 +352,7 @@ class Solution {
```
### Python
**递归法 - 前序遍历**
(版本一) 递归 - 前序 - 修改root1
```python
# Definition for a binary tree node.
# class TreeNode:
@ -377,8 +376,33 @@ class Solution:
return root1 # ⚠️ 注意: 本题我们重复使用了题目给出的节点而不是创建新节点. 节省时间, 空间.
```
(版本二) 递归 - 前序 - 新建root
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
# 递归终止条件:
# 但凡有一个节点为空, 就立刻返回另外一个. 如果另外一个也为None就直接返回None.
if not root1:
return root2
if not root2:
return root1
# 上面的递归终止条件保证了代码执行到这里root1, root2都非空.
root = TreeNode() # 创建新节点
root.val += root1.val + root2.val# 中
root.left = self.mergeTrees(root1.left, root2.left) #左
root.right = self.mergeTrees(root1.right, root2.right) # 右
**迭代法**
return root # ⚠️ 注意: 本题我们创建了新节点.
```
(版本三) 迭代
```python
class Solution:
def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
@ -413,7 +437,44 @@ class Solution:
return root1
```
(版本四) 迭代 + 代码优化
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
from collections import deque
class Solution:
def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode:
if not root1:
return root2
if not root2:
return root1
queue = deque()
queue.append((root1, root2))
while queue:
node1, node2 = queue.popleft()
node1.val += node2.val
if node1.left and node2.left:
queue.append((node1.left, node2.left))
elif not node1.left:
node1.left = node2.left
if node1.right and node2.right:
queue.append((node1.right, node2.right))
elif not node1.right:
node1.right = node2.right
return root1
```
### Go
```go

33
problems/0647.回文子串.md
View File

@ -238,33 +238,24 @@ Java：
```java
class Solution {
public int countSubstrings(String s) {
int len, ans = 0;
if (s == null || (len = s.length()) < 1) return 0;
//dp[i][j]：s字符串下标i到下标j的字串是否是一个回文串，即s[i, j]
char[] chars = s.toCharArray();
int len = chars.length;
boolean[][] dp = new boolean[len][len];
for (int j = 0; j < len; j++) {
for (int i = 0; i <= j; i++) {
//当两端字母一样时，才可以两端收缩进一步判断
if (s.charAt(i) == s.charAt(j)) {
//i++，j--，即两端收缩之后i,j指针指向同一个字符或者i超过j了,必然是一个回文串
if (j - i < 3) {
int result = 0;
for (int i = len - 1; i >= 0; i--) {
for (int j = i; j < len; j++) {
if (chars[i] == chars[j]) {
if (j - i <= 1) { // 情况一 和 情况二
result++;
dp[i][j] = true;
} else if (dp[i + 1][j - 1]) { //情况三
result++;
dp[i][j] = true;
} else {
//否则通过收缩之后的字串判断
dp[i][j] = dp[i + 1][j - 1];
}
} else {//两端字符不一样，不是回文串
dp[i][j] = false;
}
}
}
//遍历每一个字串，统计回文串个数
for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
if (dp[i][j]) ans++;
}
}
return ans;
return result;
}
}

89
problems/0654.最大二叉树.md
View File

@ -259,52 +259,75 @@ class Solution {
```
### Python
(版本一) 基础版
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
"""递归法 更快"""
def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode:
if not nums:
return None
maxvalue = max(nums)
index = nums.index(maxvalue)
if len(nums) == 1:
return TreeNode(nums[0])
node = TreeNode(0)
# 找到数组中最大的值和对应的下标
maxValue = 0
maxValueIndex = 0
for i in range(len(nums)):
if nums[i] > maxValue:
maxValue = nums[i]
maxValueIndex = i
node.val = maxValue
# 最大值所在的下标左区间 构造左子树
if maxValueIndex > 0:
new_list = nums[:maxValueIndex]
node.left = self.constructMaximumBinaryTree(new_list)
# 最大值所在的下标右区间 构造右子树
if maxValueIndex < len(nums) - 1:
new_list = nums[maxValueIndex+1:]
node.right = self.constructMaximumBinaryTree(new_list)
return node
root = TreeNode(maxvalue)
left = nums[:index]
right = nums[index + 1:]
root.left = self.constructMaximumBinaryTree(left)
root.right = self.constructMaximumBinaryTree(right)
return root
```
(版本二) 使用下标
```python
class Solution:
"""最大二叉树 递归法"""
def traversal(self, nums: List[int], left: int, right: int) -> TreeNode:
if left >= right:
return None
maxValueIndex = left
for i in range(left + 1, right):
if nums[i] > nums[maxValueIndex]:
maxValueIndex = i
root = TreeNode(nums[maxValueIndex])
root.left = self.traversal(nums, left, maxValueIndex)
root.right = self.traversal(nums, maxValueIndex + 1, right)
return root
def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode:
return self.traversal(nums, 0, len(nums))
def traversal(self, nums: List[int], begin: int, end: int) -> TreeNode:
# 列表长度为0时返回空节点
if begin == end:
```
(版本三) 使用切片
```python
class Solution:
def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode:
if not nums:
return None
max_val = max(nums)
max_index = nums.index(max_val)
node = TreeNode(max_val)
node.left = self.constructMaximumBinaryTree(nums[:max_index])
node.right = self.constructMaximumBinaryTree(nums[max_index+1:])
return node
# 找到最大的值和其对应的下标
max_index = begin
for i in range(begin, end):
if nums[i] > nums[max_index]:
max_index = i
# 构建当前节点
root = TreeNode(nums[max_index])
# 递归构建左右子树
root.left = self.traversal(nums, begin, max_index)
root.right = self.traversal(nums, max_index + 1, end)
return root
```
### Go

144
problems/0669.修剪二叉搜索树.md
View File

@ -248,6 +248,8 @@ public:
## Java
**递归**
```Java
class Solution {
public TreeNode trimBST(TreeNode root, int low, int high) {
@ -269,66 +271,114 @@ class Solution {
```
## Python
**递归**
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def trimBST(self, root: TreeNode, low: int, high: int) -> TreeNode:
'''
确认递归函数参数以及返回值：返回更新后剪枝后的当前root节点
'''
# Base Case
if not root: return None
# 单层递归逻辑
if root.val < low:
# 若当前root节点小于左界：只考虑其右子树，用于替代更新后的其本身，抛弃其左子树整体
return self.trimBST(root.right, low, high)
if high < root.val:
# 若当前root节点大于右界：只考虑其左子树，用于替代更新后的其本身，抛弃其右子树整体
return self.trimBST(root.left, low, high)
if low <= root.val <= high:
root.left = self.trimBST(root.left, low, high)
root.right = self.trimBST(root.right, low, high)
# 返回更新后的剪枝过的当前节点root
return root
```
**迭代**
```Java
class Solution {
//iteration
public TreeNode trimBST(TreeNode root, int low, int high) {
if(root == null)
return null;
while(root != null && (root.val < low || root.val > high)){
if(root.val < low)
root = root.right;
else
root = root.left;
}
TreeNode curr = root;
//deal with root's left sub-tree, and deal with the value smaller than low.
while(curr != null){
while(curr.left != null && curr.left.val < low){
curr.left = curr.left.right;
}
curr = curr.left;
}
//go back to root;
curr = root;
//deal with root's righg sub-tree, and deal with the value bigger than high.
while(curr != null){
while(curr.right != null && curr.right.val > high){
curr.right = curr.right.left;
}
curr = curr.right;
}
return root;
}
}
````
## Python
递归法（版本一）
```python
class Solution:
def trimBST(self, root: Optional[TreeNode], low: int, high: int) -> Optional[TreeNode]:
if not root: return root
# 处理头结点，让root移动到[L, R] 范围内，注意是左闭右开
while root and (root.val < low or root.val > high):
if root.val < low: # 小于L往右走
root = root.right
else: # 大于R往左走
root = root.left
# 此时root已经在[L, R] 范围内，处理左孩子元素小于L的情况
def trimBST(self, root: TreeNode, low: int, high: int) -> TreeNode:
if root is None:
return None
if root.val < low:
# 寻找符合区间 [low, high] 的节点
return self.trimBST(root.right, low, high)
if root.val > high:
# 寻找符合区间 [low, high] 的节点
return self.trimBST(root.left, low, high)
root.left = self.trimBST(root.left, low, high) # root.left 接入符合条件的左孩子
root.right = self.trimBST(root.right, low, high) # root.right 接入符合条件的右孩子
return root
```
递归法（版本二）精简
```python
class Solution:
def trimBST(self, root: TreeNode, low: int, high: int) -> TreeNode:
if root is None:
return None
if root.val < low:
return self.trimBST(root.right, low, high)
if root.val > high:
return self.trimBST(root.left, low, high)
root.left = self.trimBST(root.left, low, high)
root.right = self.trimBST(root.right, low, high)
return root
```
迭代法
```python
class Solution:
def trimBST(self, root: TreeNode, L: int, R: int) -> TreeNode:
if not root:
return None
# 处理头结点，让root移动到[L, R] 范围内，注意是左闭右闭
while root and (root.val < L or root.val > R):
if root.val < L:
root = root.right # 小于L往右走
else:
root = root.left # 大于R往左走
cur = root
# 此时root已经在[L, R] 范围内，处理左孩子元素小于L的情况
while cur:
while cur.left and cur.left.val < low:
while cur.left and cur.left.val < L:
cur.left = cur.left.right
cur = cur.left
# 此时root已经在[L, R] 范围内，处理右孩子大于R的情况
cur = root
# 此时root已经在[L, R] 范围内，处理右孩子大于R的情况
while cur:
while cur.right and cur.right.val > high:
while cur.right and cur.right.val > R:
cur.right = cur.right.left
cur = cur.right
return root
```
## Go

6
problems/0700.二叉搜索树中的搜索.md
View File

@ -230,7 +230,7 @@ class Solution {
### Python
递归法：
（方法一） 递归
```python
class Solution:
@ -250,12 +250,12 @@ class Solution:
```
迭代法：
（方法二）迭代
```python
class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
while root is not None:
while root:
if val < root.val: root = root.left
elif val > root.val: root = root.right
else: return root

183
problems/0701.二叉搜索树中的插入操作.md
View File

@ -256,133 +256,104 @@ class Solution {
-----
## Python
**递归法** - 有返回值
递归法（版本一）
```python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
# 返回更新后的以当前root为根节点的新树，方便用于更新上一层的父子节点关系链
def __init__(self):
self.parent = None
# Base Case
if not root: return TreeNode(val)
def traversal(self, cur, val):
if cur is None:
node = TreeNode(val)
if val > self.parent.val:
self.parent.right = node
else:
self.parent.left = node
return
# 单层递归逻辑:
if val < root.val:
# 将val插入至当前root的左子树中合适的位置
# 并更新当前root的左子树为包含目标val的新左子树
root.left = self.insertIntoBST(root.left, val)
self.parent = cur
if cur.val > val:
self.traversal(cur.left, val)
if cur.val < val:
self.traversal(cur.right, val)
if root.val < val:
# 将val插入至当前root的右子树中合适的位置
# 并更新当前root的右子树为包含目标val的新右子树
root.right = self.insertIntoBST(root.right, val)
# 返回更新后的以当前root为根节点的新树
def insertIntoBST(self, root, val):
self.parent = TreeNode(0)
if root is None:
return TreeNode(val)
self.traversal(root, val)
return root
```
**递归法** - 无返回值
递归法（版本二）
```python
class Solution:
def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
if not root:
def insertIntoBST(self, root, val):
if root is None:
return TreeNode(val)
parent = None
def __traverse(cur: TreeNode, val: int) -> None:
# 在函数运行的同时把新节点插入到该被插入的地方.
nonlocal parent
if not cur:
new_node = TreeNode(val)
if parent.val < val:
parent.right = new_node
else:
parent.left = new_node
return
parent = cur # 重点: parent的作用只有运行到上面if not cur:才会发挥出来.
if cur.val < val:
__traverse(cur.right, val)
else:
__traverse(cur.left, val)
return
__traverse(root, val)
return root
```
**递归法** - 无返回值 - another easier way
```python
class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
newNode = TreeNode(val)
if not root: return newNode
if not root.left and val < root.val:
root.left = newNode
if not root.right and val > root.val:
root.right = newNode
if val < root.val:
self.insertIntoBST(root.left, val)
if val > root.val:
self.insertIntoBST(root.right, val)
return root
```
**递归法** - 无返回值 有注释 不用Helper function
```python
class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if not root: # for root==None
return TreeNode(val)
if root.val<val:
if root.right==None: # find the parent
root.right = TreeNode(val)
else: # not found, keep searching
self.insertIntoBST(root.right, val)
if root.val>val:
if root.left==None: # found the parent
root.left = TreeNode(val)
else: # not found, keep searching
self.insertIntoBST(root.left, val)
# return the final tree
return root
```
**迭代法**
与无返回值的递归函数的思路大体一致
```python
class Solution:
def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
if not root:
return TreeNode(val)
parent = None # 此步可以省略
cur = root
# 用while循环不断地找新节点的parent
while cur:
parent = cur # 首先保存当前非空节点作为下一次迭代的父节点
if cur.val < val:
cur = cur.right
elif cur.val > val:
parent = cur
if val < cur.val:
cur = cur.left
# 运行到这意味着已经跳出上面的while循环,
# 同时意味着新节点的parent已经被找到.
# parent已被找到, 新节点已经ready. 把两个节点黏在一起就好了.
if parent.val > val:
else:
cur = cur.right
if val < parent.val:
parent.left = TreeNode(val)
else:
parent.right = TreeNode(val)
return root
```
递归法（版本三）
```python
class Solution:
def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
if root is None or root.val == val:
return TreeNode(val)
elif root.val > val:
if root.left is None:
root.left = TreeNode(val)
else:
self.insertIntoBST(root.left, val)
elif root.val < val:
if root.right is None:
root.right = TreeNode(val)
else:
self.insertIntoBST(root.right, val)
return root
```
迭代法
```python
class Solution:
def insertIntoBST(self, root, val):
if root is None: # 如果根节点为空，创建新节点作为根节点并返回
node = TreeNode(val)
return node
cur = root
parent = root # 记录上一个节点，用于连接新节点
while cur is not None:
parent = cur
if cur.val > val:
cur = cur.left
else:
cur = cur.right
node = TreeNode(val)
if val < parent.val:
parent.left = node # 将新节点连接到父节点的左子树
else:
parent.right = node # 将新节点连接到父节点的右子树
return root
```
-----
## Go

9
problems/二叉树理论基础.md
View File

@ -197,6 +197,7 @@ public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
@ -212,10 +213,10 @@ Python：
```python
class TreeNode:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def __init__(self, val, left = None, right = None):
self.val = val
self.left = left
self.right = right
```
Go：