LeetCode OJ：Balanced Binary Tree（平衡二叉树）

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.

就是去查看一棵树是不是平衡的，一开始对平衡二叉树的理解有错误，所以写错了 ，看了别人的解答之后更正过来了：

 /**
  * 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;
         checkBalance(root, dep);
     }
     bool checkBalance(TreeNode * root, int &dep)
     {
         if(root == NULL){
             dep = ;
             return true;
         }
         int leftDep, rightDep;
         bool isLeftBal = checkBalance(root->left, leftDep);
         bool isRightBal = checkBalance(root->right, rightDep);

         dep = max(leftDep, rightDep) + ;
         return isLeftBal && isRightBal && (abs(leftDep - rightDep) <= );
     }
 };

pS：感觉这个不应该easy的题目啊  想的时候头还挺疼的。。

用java的时候用上面的方法去做总是无法成功，所以换了一种方法，这个一开始没有想到，是看别人写的，代码如下所示：

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 public class Solution {
     public boolean isBalanced(TreeNode root) {
         if(root == null)
             return true;
         if(root.left == null && root.right == null)
             return true;
         if(Math.abs(getDep(root.left) - getDep(root.right)) > 1)
             return false;
         return isBalanced(root.left) && isBalanced(root.right);
     }

     public int getDep(TreeNode node){
         if(node == null)
             return 0;
         else
             return 1 + Math.max(getDep(node.left), getDep(node.right));
     }
 }