Problem URL: Count univalue subtrees

We will use postorder traversal to count the number of univalue subtrees. We will also use a helper function to check if the current node is a univalue subtree.

# 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 countUnivalSubtrees(self, root: Optional[TreeNode]) -> int:
        self.count = 0

        def dfs(root, parent=None):
            if not root:
                return True

            left = dfs(root.left, root.val)
            right = dfs(root.right, root.val)
            if left and right:
                self.count += 1

            return left and right and root.val == parent

        dfs(root)
        return self.count

Time complexity: O(n), n is the number of nodes in the tree
Space complexity: O(n)