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Merge branch 'youngyangyang04:master' into master

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xsduan98
2021-08-01 11:12:28 +08:00
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105
problems/0112.路径总和.md
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@ -416,6 +416,8 @@ class Solution {
Python： Python：
0112.路径总和 0112.路径总和
**递归**
```python ```python
# Definition for a binary tree node. # Definition for a binary tree node.
# class TreeNode: # class TreeNode:
@ -424,28 +426,56 @@ Python：
# self.left = left # self.left = left
# self.right = right # self.right = right
// 递归法
class Solution: class Solution:
def hasPathSum(self, root: TreeNode, targetSum: int) -> bool: def hasPathSum(self, root: TreeNode, targetSum: int) -> bool:
def isornot(root, targetSum) -> bool: def isornot(root, targetSum) -> bool:
if (not root.left) and (not root.right) and targetSum == 0:return True // 遇到叶子节点，并且计数为0 if (not root.left) and (not root.right) and targetSum == 0:
if (not root.left) and (not root.right):return False //遇到叶子节点，计数不为0 return True # 遇到叶子节点，并且计数为0
if (not root.left) and (not root.right):
return False # 遇到叶子节点，计数不为0
if root.left: if root.left:
targetSum -= root.left.val //左节点 targetSum -= root.left.val # 左节点
if isornot(root.left,targetSum):return True //递归，处理左节点 if isornot(root.left, targetSum): return True # 递归，处理左节点
targetSum += root.left.val //回溯 targetSum += root.left.val # 回溯
if root.right: if root.right:
targetSum -= root.right.val //右节点 targetSum -= root.right.val # 右节点
if isornot(root.right,targetSum):return True //递归，处理右节点 if isornot(root.right, targetSum): return True # 递归，处理右节点
targetSum += root.right.val //回溯 targetSum += root.right.val # 回溯
return False return False
if root == None:return False //别忘记处理空TreeNode if root == None:
else:return isornot(root,targetSum-root.val) return False # 别忘记处理空TreeNode
else:
return isornot(root, targetSum - root.val)
``` ```
**迭代 - 层序遍历**
```python
class Solution:
def hasPathSum(self, root: TreeNode, targetSum: int) -> bool:
if not root:
return False
stack = [] # [(当前节点，路径数值), ...]
stack.append((root, root.val))
while stack:
cur_node, path_sum = stack.pop()
if not cur_node.left and not cur_node.right and path_sum == targetSum:
return True
if cur_node.right:
stack.append((cur_node.right, path_sum + cur_node.right.val))
if cur_node.left:
stack.append((cur_node.left, path_sum + cur_node.left.val))
return False
```
0113.路径总和-ii 0113.路径总和-ii
**递归**
```python ```python
# Definition for a binary tree node. # Definition for a binary tree node.
# class TreeNode: # class TreeNode:
@ -453,35 +483,36 @@ class Solution:
# self.val = val # self.val = val
# self.left = left # self.left = left
# self.right = right # self.right = right
//递归法
class Solution: class Solution:
def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]: def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
path=[]
res=[] def traversal(cur_node, remain):
def pathes(root,targetSum): if not cur_node.left and not cur_node.right and remain == 0:
if (not root.left) and (not root.right) and targetSum == 0: // 遇到叶子节点，并且计数为0 result.append(path[:])
res.append(path[:]) //找到一种路径，记录到res中，注意必须是path[:]而不是path
return
if (not root.left) and (not root.right):return // 遇到叶子节点直接返回
if root.left: //左
targetSum -= root.left.val
path.append(root.left.val) //递归前记录节点
pathes(root.left,targetSum) //递归
targetSum += root.left.val //回溯
path.pop() //回溯
if root.right: //右
targetSum -= root.right.val
path.append(root.right.val) //递归前记录节点
pathes(root.right,targetSum) //递归
targetSum += root.right.val //回溯
path.pop() //回溯
return return
if root == None:return [] //处理空TreeNode if not cur_node.left and not cur_node.right: return
else:
path.append(root.val) //首先处理根节点 if cur_node.left:
pathes(root,targetSum-root.val) path.append(cur_node.left.val)
return res remain -= cur_node.left.val
traversal(cur_node.left, remain)
path.pop()
remain += cur_node.left.val
if cur_node.right:
path.append(cur_node.right.val)
remain -= cur_node.right.val
traversal(cur_node.right, remain)
path.pop()
remain += cur_node.right.val
result, path = [], []
if not root:
return []
path.append(root.val)
traversal(root, targetSum - root.val)
return result
``` ```
Go： Go：

56
problems/0404.左叶子之和.md
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@ -205,25 +205,51 @@ class Solution {
Python： Python：
```Python
**递归** **递归**
# Definition for a binary tree node. ```python
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution: class Solution:
def sumOfLeftLeaves(self, root: TreeNode) -> int: def sumOfLeftLeaves(self, root: TreeNode) -> int:
self.res=0 if not root:
def areleftleaves(root): return 0
if not root:return
if root.left and (not root.left.left) and (not root.left.right):self.res+=root.left.val left_left_leaves_sum = self.sumOfLeftLeaves(root.left) # 左
areleftleaves(root.left) right_left_leaves_sum = self.sumOfLeftLeaves(root.right) # 右
areleftleaves(root.right)
areleftleaves(root) cur_left_leaf_val = 0
return self.res if root.left and not root.left.left and not root.left.right:
cur_left_leaf_val = root.left.val # 中
return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum
``` ```
**迭代**
```python
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
```
Go： Go：
> 递归法 > 递归法

60
problems/0513.找树左下角的值.md
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@ -274,27 +274,51 @@ class Solution {
Python： Python：
**递归 - 回溯**
```python ```python
//递归法
# 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: class Solution:
def findBottomLeftValue(self, root: TreeNode) -> int: def findBottomLeftValue(self, root: TreeNode) -> int:
depth=0 max_depth = -float("INF")
self.res=[] leftmost_val = 0
def level(root,depth):
if not root:return def __traverse(root, cur_depth):
if depth==len(self.res): nonlocal max_depth, leftmost_val
self.res.append([]) if not root.left and not root.right:
self.res[depth].append(root.val) if cur_depth > max_depth:
level(root.left,depth+1) max_depth = cur_depth
level(root.right,depth+1) leftmost_val = root.val
level(root,depth) if root.left:
return self.res[-1][0] cur_depth += 1
__traverse(root.left, cur_depth)
cur_depth -= 1
if root.right:
cur_depth += 1
__traverse(root.right, cur_depth)
cur_depth -= 1
__traverse(root, 0)
return leftmost_val
```
**迭代 - 层序遍历**
```python
class Solution:
def findBottomLeftValue(self, root: TreeNode) -> int:
queue = deque()
if root:
queue.append(root)
result = 0
while queue:
q_len = len(queue)
for i in range(q_len):
if i == 0:
result = queue[i].val
cur = queue.popleft()
if cur.left:
queue.append(cur.left)
if cur.right:
queue.append(cur.right)
return result
``` ```
Go： Go：

27
problems/0718.最长重复子数组.md
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@ -252,6 +252,33 @@ func findLength(A []int, B []int) int {
} }
``` ```
JavaScript：
> 动态规划
```javascript
const findLength = (A, B) => {
// A、B数组的长度
const [m, n] = [A.length, B.length];
// dp数组初始化，都初始化为0
const dp = new Array(m + 1).fill(0).map(x => new Array(n + 1).fill(0));
// 初始化最大长度为0
let res = 0;
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
// 遇到A[i - 1] === B[j - 1]，则更新dp数组
if (A[i - 1] === B[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
}
// 更新res
res = dp[i][j] > res ? dp[i][j] : res;
}
}
// 遍历完成，返回res
return res;
};
```
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