[LeetCode] Subtree of Another Tree

题目链接：点击打开链接

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

     3
    / \
   4   5
  / \
 1   2

Given tree t:

   4 
  / \
 1   2

Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

     3
    / \
   4   5
  / \
 1   2
    /
   0

Given tree t:

   4
  / \
 1   2

Return false.

不得不说关于树的操作我真的太不熟悉了，以至于读完题目我都没什么想法。一般关于树的遍历必定是递归，所以我就先撸了一个递归遍历s树中的子树，对于每一棵子树都跟t树递归比较。整理了一下递归函数就可以运行了。试探性地交一发，居然没有TLE，还超过了78.67%的提交......

果然不能把LeetCode当成ACM来打= =

代码如下：

/**
 * 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 cmp(TreeNode* s, TreeNode* t)
    {
        if(s==NULL&&t==NULL)return true;
        if(s==NULL||t==NULL)return false;
        if(s->val==t->val)
        {
            return cmp(s->left,t->left)&&cmp(s->right,t->right);
        }
        else return false;
    }
    bool isSubtree(TreeNode* s, TreeNode* t) {
        if(s==NULL&&t==NULL)return true;
        if(s==NULL||t==NULL)return false;
        if(cmp(s,t))return true;
        else return isSubtree(s->left,t)||isSubtree(s->right,t);
    }
};