力扣（2024.08.10）

1. 222：完全二叉树的节点个数

# 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 countNodes(self, root: Optional[TreeNode]) -> int:def dfs(node, res):if not node:returnres.append(node.val)dfs(node.left, res)dfs(node.right, res)res = []dfs(root, res)return len(res)

 2. 404：左叶子之和

# 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: Optional[TreeNode]) -> int:def dfs(node, res, isleft):if not node:returnif isleft and not node.left and not node.right:res.append(node.val)dfs(node.left, res, True)dfs(node.right, res, False)res = []isleft = Falsedfs(root, res, isleft)return sum(res)

3. 513：找树左下角的值

# 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 findBottomLeftValue(self, root: Optional[TreeNode]) -> int:def dfs(node, level, res):if not node:returnif len(res) == level:res.append([])res[level].append(node.val)dfs(node.left, level + 1, res)dfs(node.right, level + 1, res)res = []level = 0dfs(root, level, res)return res[-1][0]

4. 112：路径总和

# 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 hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:def dfs(node, path, res):if not node:path.append('None')returnpath.append(node.val)if not node.left and not node.right:res.append(sum(path))dfs(node.left, path, res)path.pop()dfs(node.right, path, res)path.pop()res = []path = []dfs(root, path, res)if targetSum in res:return Truereturn False

 5. 113：路径总和2

# 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 pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:def dfs(node, path, res):if not node:path.append("None")returnpath.append(node.val)if not node.left and not node.right:res.append(path.copy())dfs(node.left, path, res)path.pop()dfs(node.right, path, res)path.pop()res = []path = []dfs(root, path, res)final_res = []for i in res:if sum(i) == targetSum:final_res.append(i)return final_res

6. 700：二叉搜索树中的搜索

# 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 searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:def dfs(node, val, res):if not node:returnif node.val == val:res.append(node)returnelif node.val > val:dfs(node.left, val, res)elif node.val < val:dfs(node.right, val, res)res = []dfs(root, val, res)return res[0] if res else None