[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);
}
};