LeetCode： 111. Minimum Depth of Binary Tree

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 = Noneclass Solution:def isBalanced(self, root: TreeNode) -> bool:if root is None:return Truel = self.maxDepth(root.left)r = self.maxDepth(root.right)print(l,r)if abs(l-r)>1:return Falseelse:return self.isBalanced(root.left) and self.isBalanced(root.right)def maxDepth(self, root: TreeNode) -> int:res = 0if root is None:return reslevel = [root]while level:res += 1internode = []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 = Trueself.getHeight(root)return self.isBalanceddef getHeight(self, root):if not root: return 0left = self.getHeight(root.left)right = self.getHeight(root.right)if abs(left - right) > 1: self.isBalanced = Falsereturn max(left, right) + 1

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LeetCode： 111. Minimum Depth of Binary Tree

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