Leetcode: Balanced Binary Tree

最新推荐文章于 2019-09-19 14:07:12 发布

ayst123

最新推荐文章于 2019-09-19 14:07:12 发布

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分类专栏： leetcode

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本文介绍了一种高效判断二叉树是否为高度平衡的方法。通过递归搜索每个节点的深度，并设置标记来判断子树是否平衡。若子树不平衡，则返回特殊值-1；否则返回该子树根节点的深度。

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Get idea from Code Ganker′s Solution

Question

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.

Hide Tags Tree Depth-first Search

Analysis

We need to return the depth of each node and whether it is balanced.

The tricky here is to set -1 if subtree is not balanced, otherwise return depth of root of subtree.

Other′s Solution

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

class Solution:
    # @param {TreeNode} root
    # @return {boolean}
    def isBalanced(self, root):
        return self.search(root)>=0

    def search(self,root):
        if root==None:
            return 0

        hl = self.search(root.left)
        hr = self.search(root.right)
        if hl>=0 and hr>=0 and abs(hl-hr)<=1:
            return 1 + max(hl,hr)
        else:
            return -1