> For the complete documentation index, see [llms.txt](https://luj.gitbook.io/code/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://luj.gitbook.io/code/tree/3-double-preorder/subtree-of-another-tree.md).

# Subtree of Another Tree

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**.

## Note

采用前序遍历寻找。找到父节点相同时，开始检查是不是相同的树。若不是，继续查找

## Code

```java
/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param s: the s' root
     * @param t: the t' root
     * @return: whether tree t has exactly the same structure and node values with a subtree of s
     */
    public boolean isSubtree(TreeNode s, TreeNode t) {
        // Write your code here
        if (s == null) {
            return t == null;
        }

        if (s.val == t.val && isSametree(s, t)){
            return true;
        }

        return isSubtree(s.left, t) || isSubtree(s.right, t);
    }

    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) return true;
        if (p == null || q == null) return false;
        if (p.val != q.val) return false;

        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}
```