balcanced-binary-tree

平衡二叉树判断

weixin_30740581

于 2017-09-11 21:21:00 发布

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原文链接：http://www.cnblogs.com/dd2hm/p/7507183.html

本文介绍了一种判断二叉树是否为高度平衡的方法。通过计算每个节点左右子树的深度差来判断整个二叉树是否满足高度平衡条件，即任意节点的左右子树深度之差不超过1。

题目描述

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 ofevery node never differ by more than 1.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int dept(TreeNode *root) {
        if(root==NULL)	return 0;
        if(root->left==NULL && root->right==NULL) 
            return 1;
        return max(dept(root->left),dept(root->right))+1;
    }
    
    bool isBalanced(TreeNode *root) {
        if(root==NULL)	return true;
        int x=dept(root->left)-dept(root->right);
        if(abs(x)>1)
            return false;
        else
            return isBalanced(root->left) && isBalanced(root->right);
    }
};

转载于:https://www.cnblogs.com/dd2hm/p/7507183.html