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.
给定两棵非空二叉树s和t，判定t是否为s的一棵子树。
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.
思路
递归思想，本节点不满足时则取两个子节点判定结果的逻辑或，本节点满足时则判定两个子节点结果的逻辑与。
/**
* 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 isSubtree(TreeNode* s, TreeNode* t) {
if(s==nullptr) return false;
if(isSame(s,t)) return true;
return isSubtree(s->left,t) || isSubtree(s->right,t);
}
bool isSame(TreeNode* s, TreeNode* t){
if(s==nullptr && t==nullptr) return true;
if(s==nullptr || t==nullptr) return false;
if(s->val != t->val) return false;
return isSame(s->left,t->left) && isSame(s->right,t->right);
}
};
