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));

     }

 }

巴特西

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

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