LeetCode110 Balanced Binary Tree

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详细见：leetcode.com/problems/balanced-binary-tree

Java Solution: github

package leetcode;import tools.TreeNode辅助.TreeNode;/* * 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. */public class P110_BalancedBinaryTree {public static void main(String[] args) {}/* * AC * 1 ms */static class Solution {boolean isFalse = false;    public boolean isBalanced(TreeNode root) {    if (root == null) {    return true;    }    zxwtry_count(root);        return ! isFalse;    }    int zxwtry_count(TreeNode root) {    if (isFalse) {    return 0;    }    if (root == null) {    return 0;    }    int left = zxwtry_count(root.left);    int right = zxwtry_count(root.right);    if (Math.abs(left - right) > 1) {    isFalse = true;    }    return Math.max(left, right) + 1;    }}}

C Solution: github

/*    url: leetcode.com/problems/balanced-binary-tree    AC 6ms 54.31%*/#include <stdio.h>#include <stdlib.h>struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;};#define bool inttypedef struct TreeNode stn;typedef struct TreeNode * ptn;int search(ptn root, int* b) {    int cut = 0, lc = 0, rc = 0;    if (root == NULL) return 0;    lc = search(root->left, b);    if (! (*b)) return 0;    rc = search(root->right, b);    cut = lc - rc;    if (cut < -1 || cut > 1) *b = 0;    if (! (*b)) return 0;    return lc > rc ? lc+1 : rc+1;}bool isBalanced(ptn root) {    int cut = 0, lc = 0, rc = 0, b = 1;    if (root == NULL) return 1;    lc = search(root->left, &b);    if (! b) return 0;    rc = search(root->right, &b);    if (! b) return 0;    cut = lc - rc;    return cut >= -1 && cut <= 1;}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/balanced-binary-tree    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年4月29日    @details:    Solution: 82ms 70.51%'''class TreeNode(object):    def __init__(self, x):        self.val = x        self.left = None        self.right = Noneclass Solution(object):    def search(self, n, sign):        if n == None or sign[0]: return 0        lc = self.search(n.left, sign)        rc = self.search(n.right, sign)        if abs(lc-rc) > 1: sign[0] = True        return max(lc, rc) + 1            def isBalanced(self, n):        """        :type n: TreeNode        :rtype: bool        """        if n == None: return True        sign = [False]        lc = self.search(n.left, sign)        if (sign[0]): return False        rc = self.search(n.right, sign)        if (sign[0]): return False        return abs(lc-rc) < 2

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