How do you solve Diameter of Binary Tree on LeetCode?

Diameter of Binary Tree (LeetCode #543) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: The diameter can be calculated without explicitly finding all pairs of nodes, making the optimal solution much more efficient.

What is the brute force solution for Diameter of Binary Tree?

The brute force approach for Diameter of Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves calculating the diameter by checking all possible pairs of nodes in the tree. For each pair, we find the path between them and count the edges, ultimately returning the maximum path length found. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Diameter of Binary Tree?

Diameter of Binary Tree uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Tree Traversal.

What are the interview tips for Diameter of Binary Tree?

Always clarify the definition of diameter with the interviewer to ensure you're solving the right problem. Discuss your thought process and approach before jumping into coding; this shows your understanding. Practice explaining your code as you write it, as this simulates the interview environment.

What are common mistakes when solving Diameter of Binary Tree?

Students often forget that the diameter does not necessarily pass through the root, leading to incorrect calculations. Not keeping track of the maximum diameter during the DFS traversal can result in missing the longest path.

#543

Diameter of Binary Tree

Easy

TreeDepth-First SearchBinary TreeDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(h)

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Intuition

Time O(n)Space O(h)

The optimal solution uses a single depth-first search (DFS) to calculate the diameter in one pass. By keeping track of the maximum diameter during the traversal, we avoid redundant calculations and achieve better efficiency.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the maximum diameter.
2Step 2: Perform a DFS on the tree, calculating the height of each subtree.
3Step 3: During the DFS, update the maximum diameter whenever a node's left and right heights are summed.

solution.py20 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def diameterOfBinaryTree(self, root: TreeNode) -> int:
10        def dfs(node):
11            if not node:
12                return 0
13            left = dfs(node.left)
14            right = dfs(node.right)
15            self.max_diameter = max(self.max_diameter, left + right)
16            return max(left, right) + 1
17
18        self.max_diameter = 0
19        dfs(root)
20        return self.max_diameter

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(h) due to the recursion stack, where h is the height of the tree.

1The diameter can be calculated without explicitly finding all pairs of nodes, making the optimal solution much more efficient.
2Understanding how to traverse a tree using DFS is crucial for solving many tree-related problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.