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Balanced Binary Tree

时间:2015-01-09 21:01:27      阅读:180      评论:0      收藏:0      [点我收藏+]

标签:leetcode

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.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isBalanced(TreeNode root) {
        int[] res = inorder(root);//res[0] 是否是,res[1]存储depth
        if(res[0] == 0){
            return false;
        }else{
            return true;
        }
    }
    public int[] inorder(TreeNode root){
        int res[] = new int[2];
        if(root == null){//空节点
            res[0] = 1;
            res[1] = 0;
            return res;
        }
        int[] left = inorder(root.left);
        if(left[0] == 0){//如果左子树不平衡,则没有必要继续遍历
            res[0] = 0;
            return res;
        }
        int[] right = inorder(root.right);
        if(right[0] == 0){
            res[0] = 0;
            return res;
        }
        if(Math.abs(left[1] - right[1]) <= 1){
            res[0] = 1;
        }else{
            res[0] = 0;
        }
        res[1] = (left[1] > right[1] ? left[1] : right[1]) + 1;
        return res;
    }
}

是关于树的深度问题,基本思想是DFS。不过要在DFS过程中记录其他的信息,深度和是否是平衡树这两个信息量,故设计的DFS返回的是一个数组,这样就方便处理了。

Runtime: 269 ms

Balanced Binary Tree

标签:leetcode

原文地址:http://blog.csdn.net/havedream_one/article/details/42558837

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