每日一小练——因子分解

标签：leetcode   oj   算法   

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.

分析：判断一棵树是否是平衡树，只要判断左右子树的树深差值，如果差值大于1则非，如果小于等于1则是，递归判断左右子树。

class Solution {
public:
    bool isBalanced(TreeNode *root) {
        if(!root){
            return true;
        }
        isBalancedTree = true;
        calDepth(root);
        return isBalancedTree;
    }
    int calDepth(TreeNode* root){
        if(!isBalancedTree){
            return 0;
        }
        if(!root->left && !root->right){
            return 1;
        }
        int left = 0;
        int right = 0;
        if(root->left){
            left = calDepth(root->left);
        }
        if(root->right){
            right = calDepth(root->right);
        }
        if(abs(left-right) > 1){
            isBalancedTree = false;
        }
        return max(right,left)+1;
    }
    bool isBalancedTree;
};

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每日一小练——因子分解

标签：leetcode   oj   算法   

原文地址：http://blog.csdn.net/zhurui_idea/article/details/26169133