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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.
Solution 1:
思路：首先判断root.left和root.right是否有符合条件的depth，加入辅助函数，也就是104的maxdepth，然后再递归检查左子树和右子树。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean isBalanced(TreeNode root) { if(root==null) { return true; } if(Math.abs(depth(root.left)-depth(root.right))<=1) { return isBalanced(root.left)&&isBalanced(root.right); } else { return false; } } public int depth(TreeNode root) { if(root==null) { return 0; } return Math.max(depth(root.left),depth(root.right))+1; } }
Solution 2:
discussion看到一种更巧妙的做法：
public boolean isBalanced(TreeNode root) { return getDepth(root) != -1; } private int getDepth(TreeNode root) { if (root != null) { int left = getDepth(root.left); int right = getDepth(root.right); return (left == -1 || right == -1 || Math.abs(left-right) > 1) ? -1 : Math.max(left, right) + 1; } return 0; }
Define a getDepth method, which will return the height of the tree, if it is balanced, otherwise, return -1.
定义的这个getDepth行使两个功能，一是记录depth如果balanced，二是记录－1如果不balanced。
定义好一个recursion的函数很重要啊！！
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原文地址：http://www.cnblogs.com/Machelsky/p/5858612.html
