# Solving Leetcode Interviews in Seconds with AI: Univalued Binary Tree


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "965" 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
	> A binary tree is uni-valued if every node in the tree has the same value. Given the root of a binary tree, return true if the given tree is uni-valued, or false otherwise.   Example 1:   Input: root = [1,1,1,1,1,null,1] Output: true  Example 2:   Input: root = [2,2,2,5,2] Output: false    Constraints:  The number of nodes in the tree is in the range [1, 100]. 0 <= Node.val < 100  

	# Explanation
	Here's the breakdown of the approach, complexity, and the Python code:

*   **High-Level Approach:**
    *   Use Depth-First Search (DFS) to traverse the tree.
    *   Compare the value of each node with the value of the root node.
    *   If any node's value differs from the root's value, the tree is not uni-valued.

*   **Complexity:**
    *   Runtime Complexity: O(N), where N is the number of nodes in the tree.
    *   Storage Complexity: O(H), where H is the height of the tree. In the worst case (skewed tree), H = N; in the best case (balanced tree), H = log N.

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

def isUnivalTree(root: TreeNode) -> bool:
    if not root:
        return True

    val = root.val

    def is_unival(node: TreeNode) -> bool:
        if not node:
            return True
        if node.val != val:
            return False
        return is_unival(node.left) and is_unival(node.right)

    return is_unival(root)
	```