LeetCode： 111. Minimum Depth of Binary Tree

最新推荐文章于 2023-12-15 03:22:29 发布

原创最新推荐文章于 2023-12-15 03:22:29 发布 · 108 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

34 篇文章

订阅专栏

本文介绍了一种判断二叉树是否平衡的方法，并提供了两种实现方案。平衡二叉树定义为对于任意节点，其左右子树的高度差不超过1。文中通过递归方式检查每个节点，并利用最大深度辅助函数进行高度计算。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7
Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

解题思路

这一道题是要判断二叉树是否为平衡二叉树，需要递归的判断每个节点是不是平衡的。因为需要查询树的高度，可以直接调用LeetCode: 104. Maximum Depth of Binary Tree的代码。

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

class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        if root is None:
            return True
        
        l = self.maxDepth(root.left)
        r = self.maxDepth(root.right)
        print(l,r)
        if abs(l-r)>1:
            return False
        else:
            return self.isBalanced(root.left) and self.isBalanced(root.right)
            
        
    def maxDepth(self, root: TreeNode) -> int:
        res = 0
        if root is None:
            return res
        level = [root]
        while level:
            res += 1
            internode = []
            for node in level:
                internode.extend([node.left, node.right])
            level = [leaf for leaf in internode if leaf]
        return res

LeetCode上大佬的解法：

class Solution(object):
    def isBalanced(self, root):
        self.isBalanced = True
        self.getHeight(root)
        return self.isBalanced
        
    
    def getHeight(self, root):
        if not root: return 0
        left = self.getHeight(root.left)
        right = self.getHeight(root.right)
        if abs(left - right) > 1: 
            self.isBalanced = False
        return max(left, right) + 1