> For the complete documentation index, see [llms.txt](https://dfine.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://dfine.gitbook.io/leetcode/572.subtree_of_another_tree.md).

# 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


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