leetcode 110. Balanced Binary Tree

Posted by 111qqz on Friday, February 24, 2017

TOC

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;
    }
};