Forgetting null/base-case handling
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.
Build confidence with an intuition-first walkthrough focused on tree fundamentals.
Given the root of a binary tree, return the length of the diameter of the tree.
The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
The length of a path between two nodes is represented by the number of edges between them.
Example 1:
Input: root = [1,2,3,4,5] Output: 3 Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
Example 2:
Input: root = [1,2] Output: 1
Constraints:
[1, 104].-100 <= Node.val <= 100Problem summary: Given the root of a binary tree, return the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root. The length of a path between two nodes is represented by the number of edges between them.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[1,2,3,4,5]
[1,2]
diameter-of-n-ary-tree)longest-path-with-different-adjacent-characters)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #543: Diameter of Binary Tree
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int ans;
public int diameterOfBinaryTree(TreeNode root) {
dfs(root);
return ans;
}
private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int l = dfs(root.left);
int r = dfs(root.right);
ans = Math.max(ans, l + r);
return 1 + Math.max(l, r);
}
}
// Accepted solution for LeetCode #543: Diameter of Binary Tree
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func diameterOfBinaryTree(root *TreeNode) (ans int) {
var dfs func(root *TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
l, r := dfs(root.Left), dfs(root.Right)
ans = max(ans, l+r)
return 1 + max(l, r)
}
dfs(root)
return
}
# Accepted solution for LeetCode #543: Diameter of Binary Tree
# 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 diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode]) -> int:
if root is None:
return 0
l, r = dfs(root.left), dfs(root.right)
nonlocal ans
ans = max(ans, l + r)
return 1 + max(l, r)
ans = 0
dfs(root)
return ans
// Accepted solution for LeetCode #543: Diameter of Binary Tree
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn diameter_of_binary_tree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut ans = 0;
fn dfs(root: Option<Rc<RefCell<TreeNode>>>, ans: &mut i32) -> i32 {
match root {
Some(node) => {
let node = node.borrow();
let l = dfs(node.left.clone(), ans);
let r = dfs(node.right.clone(), ans);
*ans = (*ans).max(l + r);
1 + l.max(r)
}
None => 0,
}
}
dfs(root, &mut ans);
ans
}
}
// Accepted solution for LeetCode #543: Diameter of Binary Tree
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function diameterOfBinaryTree(root: TreeNode | null): number {
let ans = 0;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
const [l, r] = [dfs(root.left), dfs(root.right)];
ans = Math.max(ans, l + r);
return 1 + Math.max(l, r);
};
dfs(root);
return ans;
}
Use this to step through a reusable interview workflow for this problem.
BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.
Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.
Review these before coding to avoid predictable interview regressions.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.