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。。。
1 /**
2 * Definition for a binary tree node.
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
10class Solution { //错误原因：左右子树都平衡的树未必平衡！！！！
11public:
12 bool leaf(TreeNode* root)
13 {
14 if (root->left==NULL&&root->right==NULL) return true;
15 return false;
16 }
17 int dep(TreeNode* root)
18 {
19 if (root==NULL) return 0;
20 return max(dep(root->left),dep(root->right))+1;
21 }
22 bool dfs(TreeNode* root)
23 {
24 bool res = true;
25 if (abs(dep(root->left)-dep(root->right))>1) return false;
26 if (root->left!=NULL) res = dfs(root->left);
27 if (!res) return false;
28 if (root->right!=NULL) res = dfs(root->right);
29 if (!res) return false;
30 return true;
31 }
32 bool isBalanced(TreeNode* root) {
33 if (root==NULL) return true;
34 bool res = dfs(root);
35 return res;
36 }
37};