Leetcode #965: Univalued Binary Tree
In this guide, we solve Leetcode #965 Univalued Binary Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root = [1,1,1,1,1,null,1]
Output: true
Python Solution
# 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:
def dfs(root: Optional[TreeNode]) -> bool:
if root is None:
return True
return root.val == x and dfs(root.left) and dfs(root.right)
x = root.val
return dfs(root)
Complexity
The time complexity is , and the space complexity is , where is the number of nodes in the tree. The space complexity is , where is the number of nodes in the tree.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.