# 572. Subtree of Another Tree

## Easy

***

Given the roots of two binary trees `root` and `subRoot`, return `true` if there is a subtree of `root` with the same structure and node values of `subRoot` and `false` otherwise.

A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/04/28/subtree1-tree.jpg)

```
Input: root = [3,4,5,1,2], subRoot = [4,1,2]
Output: true
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/04/28/subtree2-tree.jpg)

```
Input: root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
Output: false
```

**Constraints:**

* The number of nodes in the `root` tree is in the range `[1, 2000]`.
* The number of nodes in the `subRoot` tree is in the range `[1, 1000]`.
* `-104 <= root.val <= 104`
* `-104 <= subRoot.val <= 104`

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
        if subRoot is None:
            return True
        if root is None:
            return False
        if self.compare(root,subRoot):
            return True
        return self.isSubtree(root.left,subRoot) or self.isSubtree(root.right, subRoot)
    
    def compare(self, root, subRoot):
        if root is None and subRoot is None:
            return True
        if root is None or subRoot is None:
            return False
        return root.val == subRoot.val and self.compare(root.left,subRoot.left) and self.compare(root.right, subRoot.right)
        
        
```