leetcode 110. Balanced Binary Tree

题目链接

题意：判断一颗二叉树是否平衡....

思路：直接搞就好了。。。神TM又忘记dfs的时候忘记返回子调用的值。。。。我这是药丸啊。。。

 /**
 * 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 {  //错误原因：左右子树都平衡的树未必平衡！！！！
public:
		bool leaf(TreeNode* root)
		{
			if (root->left==NULL&&root->right==NULL) return true;
			return false;
		}
		int dep(TreeNode* root)
		{
			if (root==NULL) return 0;
			return max(dep(root->left),dep(root->right))+1;
		}
		bool dfs(TreeNode* root)
		{
				bool res = true;
				if (abs(dep(root->left)-dep(root->right))>1) return false;
				if (root->left!=NULL) res = dfs(root->left);
				if (!res) return false;
				if  (root->right!=NULL) res = dfs(root->right);
				if (!res) return false;
				return true;		
		}
    bool isBalanced(TreeNode* root) {
		if (root==NULL) return true;
		bool res = dfs(root);
		return res;
    }
};