【leetcode】 Balanced Binary Tree

From : https://leetcode.com/problems/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.

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isBalanced(TreeNode* root, int& dep) {if(!root) {dep=0; return true;}int left, right;if(isBalanced(root->left, left) && isBalanced(root->right, right)) {if(abs(right-left) <= 1) {dep = 1+max(left, right);return true;}}return false;}        bool isBalanced(TreeNode *root) {        int dep;        return isBalanced(root, dep);    }};

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {private static class Depth {int v = 0;}public boolean isBalanced(TreeNode root) {if (root == null) {return true;}return isBalanced(root, new Depth());}// dep is the deep of rootboolean isBalanced(TreeNode root, Depth dep) {if (root == null) {dep.v = 0;return true;}Depth left = new Depth(), right = new Depth();if (isBalanced(root.left, left) && isBalanced(root.right, right)) {if (Math.abs(right.v - left.v) <= 1) {dep.v = 1 + Math.max(right.v, left.v);return true;}}return false;}}