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题目：
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.
思路：
1) 计算结点的高度，但这个有重复计算
package tree; public class BalancedBinaryTree { public boolean isBalanced(TreeNode root) { if (root == null) return true; int leftHeight = height(root.left); int rightHeight = height(root.right); return Math.abs(leftHeight - rightHeight) <= 1 && isBalanced(root.left) && isBalanced(root.right); } private int height(TreeNode root) { if (root == null) return 0; return Math.max(1 + height(root.left), 1 + height(root.right)); } }
2) 不进行重复计算，但需要new 对象，反而也花时间
package tree; public class BalancedBinaryTree { class Entry{ public int height; public boolean balanced; } public boolean isBalanced(TreeNode root) { return checkTree(root).balanced; } private Entry checkTree(TreeNode root) { Entry entry = new Entry(); if (root == null) { entry.height = 0; entry.balanced = true; } else { Entry left = checkTree(root.left); Entry right = checkTree(root.right); entry.height = Math.max(left.height, right.height) + 1; entry.balanced = Math.abs(left.height - right.height) <= 1 && left.balanced && right.balanced; } return entry; } }
LeetCode - Balanced Binary Tree
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5118171.html
