LeetCode | Balanced Binary Tree

来源：互联网发布：淘宝评价管理截图编辑：程序博客网时间：2024/05/17 06:36

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.

OJ's Binary Tree Serialization: //LeetCode对二叉树的表示方法

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */

public class Solution {        public int maxDepth(TreeNode root){        //此方法为10.26日的题，递归求树的最大深度           if(root==null) return 0;           int leftDepth = maxDepth(root.left);           int rightDepth = maxDepth(root.right);           int depth =  leftDepth>=rightDepth?leftDepth:rightDepth ;            return depth+1;       }           public boolean isBalanced(TreeNode root) {        if (root==null) return true;               //注意，空树也认为是平衡树的情况       int leftHeight = maxDepth(root.left);       //求左右子树的高度       int rightHeight = maxDepth(root.right);             if( Math.abs(leftHeight-rightHeight)>1 ) return false;  //左右子树的高度相差大于1，一定不是平衡树       return isBalanced(root.left)&&isBalanced(root.right);   //注意！！！此处不能直接返回FALSE，要继续向下判断，    }                                                          //直至确认每个节点都满足平衡的条件才返回true    }//左右子树的高度差不大于一，并不代表一定是平衡树，平衡树要求每个节点的左右子节点高度差都不大于1，不只是root节点//例如：树{1 2 2 3 # # 3 4 # # 4}，其根的左右子树高度一样，但其不是平衡树(#代表此节点为null)

0 0