Leetcode #2773: Height of Special Binary Tree
In this guide, we solve Leetcode #2773 Height of Special 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
You are given a root, which is the root of a special binary tree with n nodes. The nodes of the special binary tree are numbered from 1 to n.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- 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 = [1,2,3,null,null,4,5]
Output: 2
Explanation: The given tree is shown in the following picture. Each leaf's left child is the leaf to its left (shown with the blue edges). Each leaf's right child is the leaf to its right (shown with the red edges). We can see that the graph has a height of 2.
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 heightOfTree(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode], d: int):
nonlocal ans
ans = max(ans, d)
if root.left and root.left.right != root:
dfs(root.left, d + 1)
if root.right and root.right.left != root:
dfs(root.right, d + 1)
ans = 0
dfs(root, 0)
return ans
Complexity
The time complexity is , and the space complexity is .
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.