leetcode刷题（五）

最新推荐文章于 2025-05-08 14:25:41 发布

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本文深入探讨了二叉树的四种核心算法：翻转二叉树、求最大深度、验证二叉查找树及路径总和。通过递归方法详细解释了每种算法的实现过程，为读者提供了清晰的代码示例和思路解析。

二叉树：

Invert Binary Tree（翻转二叉树）

英文版：https://leetcode.com/problems/invert-binary-tree/

中文版：https://leetcode-cn.com/problems/invert-binary-tree/

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return root
        root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)   # 递归
        return root

Maximum Depth of Binary Tree（二叉树的最大深度）

英文版：https://leetcode.com/problems/maximum-depth-of-binary-tree/

中文版：https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if not root:
            return 0
        else:
            l = 1 + self.maxDepth(root.left)   # 递归
            r = 1 + self.maxDepth(root.right)
            return max(l, r)

Validate Binary Search Tree（验证二叉查找树）

英文版：https://leetcode.com/problems/validate-binary-search-tree/

中文版：https://leetcode-cn.com/problems/validate-binary-search-tree/

# leetcode-98:
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isValidBST(self, root: TreeNode) -> bool:
        if not root:
            return True
        
        pre = None
        stack = list()
        while root or stack:
            if root:
                stack.append(root)
                root = root.left
            else:
                root = stack.pop()
                if pre and root.val <= pre.val:
                    return False
                
                pre = root
                root = root.right
        return True

# reference:https://blog.youkuaiyun.com/qq_17550379/article/details/82315830

Path Sum（路径总和）

英文版：https://leetcode.com/problems/path-sum/

中文版：https://leetcode-cn.com/problems/path-sum/

# leetcode-112:
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if root is None:
            return False
        if sum == root.val and root.left is None and root.right is None:
            return True
        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)   # 递归