Univalued Binary Tree

965. Univalued Binary Tree | Easy

Another one. It was an easy problem. Another challenge could be to use the least amount of code?

Problem:

A binary tree is univalued if every node in the tree has the same value.

Return true if and only if the given tree is univalued.

 

Example 1:


Input: [1,1,1,1,1,null,1]
Output: true
Example 2:


Input: [2,2,2,5,2]
Output: false
 

Note:

The number of nodes in the given tree will be in the range [1, 100].
Each node's value will be an integer in the range [0, 99].

Accepted 7,881 Submissions 10,935

Both used same code, but resulted in slightly different speed: Runtime: 56 ms, faster than 85.93% of Python3 online submissions for Univalued Binary Tree. Runtime: 52 ms, faster than 94.34% of Python3 online submissions for Univalued Binary Tree.

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isUnivalTree(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not (root.left or root.right):
            return True
        elif root.left and root.right:
            return \
                root.left.val == root.val and       \
                root.right.val == root.val and      \
                self.isUnivalTree(root.left) and    \
                self.isUnivalTree(root.right)
        elif root.left:
            return \
                root.left.val == root.val and       \
                self.isUnivalTree(root.left)
        elif root.right:
            return \
                root.right.val == root.val and      \
                self.isUnivalTree(root.right)
        return False