<p>A binary tree is <strong>uni-valued</strong> if every node in the tree has the same value.</p>

<p>Given the <code>root</code> of a binary tree, return <code>true</code><em> if the given tree is <strong>uni-valued</strong>, or </em><code>false</code><em> otherwise.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/28/unival_bst_1.png" style="width: 265px; height: 172px;" />
<pre>
<strong>Input:</strong> root = [1,1,1,1,1,null,1]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/28/unival_bst_2.png" style="width: 198px; height: 169px;" />
<pre>
<strong>Input:</strong> root = [2,2,2,5,2]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 100]</code>.</li>
	<li><code>0 &lt;= Node.val &lt; 100</code></li>
</ul>

# 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 isUnivalTree(self, root: Optional[TreeNode]) -> bool:
        if not root:
            return True
        
        def dfs(node):
            if not node:
                return True
            # Check if the current node's value is different from the root's value
            if node.val != root.val:
                return False
            # Continue to check left and right subtrees
            return dfs(node.left) and dfs(node.right)

        return dfs(root)
16500