572.Subtree of Another Tree

572.Subtree of Another Tree

难度:Easy

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.

需要子树的要求,在子树到达叶子节点的时候,原数也是叶子节点。

/**
* 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 {
bool isEqual(TreeNode*s,TreeNode*t){
if(!t) return s? false :true;
if(!s) return false;
if(s->val != t->val) return false;
return isEqual(s->left,t->left) && isEqual(s->right,t->right);
}
public:
bool isSubtree(TreeNode* s, TreeNode* t) {
if(!s && !t) return true;
if(!s || !t) return false;
if(s->val == t->val && isEqual(s,t)) return true;
return isSubtree(s->left,t) || isSubtree(s->right,t);
}
};

执行用时 :36 ms, 在所有 C++ 提交中击败了86.12%的用户 内存消耗 :21 MB, 在所有 C++ 提交中击败了77.88%的用户