Balanced Binary Tree（平衡二叉树）

来源：https://leetcode.com/problems/balanced-binary-tree

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.

平衡二叉树：是它一棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树。

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private boolean isBalancedFlag = true;
    private int getDepth(TreeNode root) {
        if(root == null) {
            return 0;
        }
        int leftDepth = getDepth(root.left) + 1;
        int rightDepth = getDepth(root.right) + 1;
        if(Math.abs(leftDepth - rightDepth) > 1) {
            isBalancedFlag = false;
        }
        return leftDepth > rightDepth ? leftDepth : rightDepth;
    }
    public boolean isBalanced(TreeNode root) {
        getDepth(root);
        return isBalancedFlag;
    }
}// 1 ms

Python

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    __is_balanced = True
    def getDepth(self, pRoot):
        if pRoot == None:
            return 0
        left_depth = self.getDepth(pRoot.left)
        right_depth = self.getDepth(pRoot.right)
        if abs(left_depth-right_depth) > 1:
            self.__is_balanced = False
        return left_depth+1 if left_depth>right_depth else right_depth+1
    def IsBalanced_Solution(self, pRoot):
        self.getDepth(pRoot)
        return self.__is_balanced