【leetcode】判断平衡二叉树

甄心爱学习2026-02-09 18:13

给定一个二叉树，判断它是否是平衡二叉树

二叉搜索树（BST）

性质：左子树所有节点值 < 根节点值 < 右子树所有节点值
目的：快速查找（O(log n) 在平衡情况下）
不保证平衡

平衡二叉树

性质：每个节点左右子树高度差不超过1
目的：避免树退化成链表，保持操作效率
不保证有序性

python 复制代码

# 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 isBalanced(self, root: Optional[TreeNode]) -> bool:
        self.res = True
        def helper(root):
            if not root:
                return 0
            left = helper(root.left) + 1
            right = helper(root.right) + 1
            #print(right, left)
            if abs(right - left) > 1: 
                self.res = False
            return max(left, right)
        helper(root)
        return self.res

python 复制代码

class Solution:
    def isBalanced(self, root: Optional[TreeNode]) -> bool:
        def height(root):
            if  not root:
                return 0
            return max(height(root.left),height(root.right))+1
        
        if not root:
            return True

        return abs(height(root.left)-height(root.right))<=1 and self.isBalanced(root.left) and self.isBalanced(root.right)