# Solving Leetcode Interviews in Seconds with AI: Subtree of Another Tree


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "572" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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:   Input: root = [3,4,5,1,2], subRoot = [4,1,2] Output: true  Example 2:   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  

	# Explanation
	*   The core idea is to traverse the main tree (`root`) and check if any of its nodes can be the root of a subtree that is identical to `subRoot`.
*   A helper function is used to recursively compare two trees for structural and value equality.
*   The traversal stops as soon as a matching subtree is found, optimizing the search.

*   **Time Complexity:** O(m*n) where n is the number of nodes in `root` and m is the number of nodes in `subRoot`. In the worst case, we might visit every node in `root` and potentially compare it against `subRoot`.
    **Space Complexity:** O(max(h1, h2)) where h1 and h2 are the heights of the two trees `root` and `subRoot` respectively, due to the recursive call stack. In the worst-case (skewed tree), this could be O(n) or O(m).

	
	# Code
	```python
	class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def isSubtree(root: TreeNode, subRoot: TreeNode) -> bool:
    """
    Checks if subRoot is a subtree of root.

    Args:
        root: The root of the main binary tree.
        subRoot: The root of the subtree to search for.

    Returns:
        True if subRoot is a subtree of root, False otherwise.
    """

    def is_identical(node1: TreeNode, node2: TreeNode) -> bool:
        """
        Recursively checks if two trees are identical in structure and value.
        """
        if not node1 and not node2:
            return True
        if not node1 or not node2 or node1.val != node2.val:
            return False
        return is_identical(node1.left, node2.left) and is_identical(node1.right, node2.right)

    def traverse(node: TreeNode, sub_root: TreeNode) -> bool:
        """
        Traverses the main tree and checks for subtree equality.
        """
        if not node:
            return False

        if is_identical(node, sub_root):
            return True

        return traverse(node.left, sub_root) or traverse(node.right, sub_root)

    return traverse(root, subRoot)
	```