diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 8fc973e0..a0bab999 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -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 ``` diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index a3bc77fb..c0561e10 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -553,6 +553,52 @@ 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 +622,6 @@ class Solution: height_map[real_node] = 1 + max(left, right) return True ``` - - ### Go ```Go diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index c396f4a0..7bd56fbd 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -470,38 +470,34 @@ 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 [] + 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 + 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 = [] - self.generate_paths(root, [], result) + path = [] + if not root: + return result + self.traversal(root, path, 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) +递归法+隐形回溯(版本一) ```Python # Definition for a binary tree node. # class TreeNode: @@ -509,7 +505,6 @@ class Solution: # self.val = val # self.left = left # self.right = right -import copy from typing import List, Optional class Solution: @@ -517,23 +512,23 @@ class Solution: if not root: return [] result = [] - self.generate_paths(root, [], result) + self.traversal(root, [], result) return result - def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None: - if not node: + def traversal(self, cur: TreeNode, path: List[int], result: List[str]) -> None: + if not cur: return - path.append(node.val) - if not node.left and not node.right: + 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() diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index 617978b7..aa2868df 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -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) - - return left_val + right_val + rightValue = self.sumOfLeftLeaves(root.right) # 右 + + 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 - - 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 + 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 + ``` ### Go diff --git a/problems/0513.找树左下角的值.md b/problems/0513.找树左下角的值.md index 26768c74..743b0df9 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -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 + ``` diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md index f98948f0..18a245c4 100644 --- a/problems/0617.合并二叉树.md +++ b/problems/0617.合并二叉树.md @@ -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 + + return root # ⚠️ 注意: 本题我们创建了新节点. -**迭代法** +``` + +(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 diff --git a/problems/0654.最大二叉树.md b/problems/0654.最大二叉树.md index 64b38b48..33a9176e 100644 --- a/problems/0654.最大二叉树.md +++ b/problems/0654.最大二叉树.md @@ -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) +``` +(版本二) 使用下标 +```python - left = nums[:index] - right = nums[index + 1:] - - root.left = self.constructMaximumBinaryTree(left) - root.right = self.constructMaximumBinaryTree(right) - return root - 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: - return None - - # 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 -``` + ``` + +(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 + +``` ### Go diff --git a/problems/0700.二叉搜索树中的搜索.md b/problems/0700.二叉搜索树中的搜索.md index 40724f48..13064f97 100644 --- a/problems/0700.二叉搜索树中的搜索.md +++ b/problems/0700.二叉搜索树中的搜索.md @@ -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