110. Balanced Binary Tree

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.

第一种递归解法

这个解法就是从根节点开始依次查各个节点的左右子节点深度，一致则从左右节点开始搜一次深度。一旦发现左右深度不一致则得到结论。该算法很有可能造成重复搜索，所以相对耗时。代码如下：

public class Solution {    public boolean isBalanced(TreeNode root) {       if(root == null) return true;       if(Math.abs(getDepth(root.left) - getDepth(root.right)) < 2) {           return isBalanced(root.left) && isBalanced(root.right);       } else return false;    }    public int getDepth(TreeNode node) {        if(node == null) return 0;        int left = getDepth(node.left);        int right = getDepth(node.right);        return Math.max(left, right) + 1;    }}

第二种递归解法

创建一个方法，参数是当前结点深度，返回值是左右节点所能到的深度。相比较上一个算法不会有重复搜索的情况，所以耗时较少。代码如下：

public class Solution {    public boolean isBalanced(TreeNode root) {        if(root == null) return true;        return judge(root, 0) > -1;    }    public int judge(TreeNode node, int depth) {        if(node == null) {            return depth - 1;        }        int left = judge(node.left, depth + 1);        int right = judge(node.right, depth + 1);        if(left != -1 && right != -1 && Math.abs(left - right) < 2) {            return Math.max(left, right);        } else {            return -1;        }    }}

同时分享一个和我这个解法思路相同，但是更加清晰简洁的代码写法。链接：
Java solution based on height, check left and right node in every recursion to avoid further useless search