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.
Move from brute-force thinking to an efficient approach using tree strategy.
Given the root of a binary tree, return the length of the longest path, where each node in the path has the same value. This path may or may not pass through the root.
The length of the path between two nodes is represented by the number of edges between them.
Example 1:
Input: root = [5,4,5,1,1,null,5] Output: 2 Explanation: The shown image shows that the longest path of the same value (i.e. 5).
Example 2:
Input: root = [1,4,5,4,4,null,5] Output: 2 Explanation: The shown image shows that the longest path of the same value (i.e. 4).
Constraints:
[0, 104].-1000 <= Node.val <= 10001000.Problem summary: Given the root of a binary tree, return the length of the longest path, where each node in the path has the same value. This path may or may not pass through the root. The length of the 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
[5,4,5,1,1,null,5]
[1,4,5,4,4,null,5]
binary-tree-maximum-path-sum)count-univalue-subtrees)path-sum-iii)longest-path-with-different-adjacent-characters)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #687: Longest Univalue Path
/**
* 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 longestUnivaluePath(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);
l = root.left != null && root.left.val == root.val ? l + 1 : 0;
r = root.right != null && root.right.val == root.val ? r + 1 : 0;
ans = Math.max(ans, l + r);
return Math.max(l, r);
}
}
// Accepted solution for LeetCode #687: Longest Univalue Path
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func longestUnivaluePath(root *TreeNode) (ans int) {
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
l, r := dfs(root.Left), dfs(root.Right)
if root.Left != nil && root.Left.Val == root.Val {
l++
} else {
l = 0
}
if root.Right != nil && root.Right.Val == root.Val {
r++
} else {
r = 0
}
ans = max(ans, l+r)
return max(l, r)
}
dfs(root)
return
}
# Accepted solution for LeetCode #687: Longest Univalue Path
# 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 longestUnivaluePath(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)
l = l + 1 if root.left and root.left.val == root.val else 0
r = r + 1 if root.right and root.right.val == root.val else 0
nonlocal ans
ans = max(ans, l + r)
return max(l, r)
ans = 0
dfs(root)
return ans
// Accepted solution for LeetCode #687: Longest Univalue Path
// 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 {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, target: i32, ans: &mut i32) -> i32 {
if root.is_none() {
return 0;
}
let root = root.as_ref().unwrap().borrow();
let left = Self::dfs(&root.left, root.val, ans);
let right = Self::dfs(&root.right, root.val, ans);
*ans = (*ans).max(left + right);
if root.val == target {
return left.max(right) + 1;
}
0
}
pub fn longest_univalue_path(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
if root.is_none() {
return 0;
}
let mut ans = 0;
Self::dfs(&root, root.as_ref().unwrap().borrow().val, &mut ans);
ans
}
}
// Accepted solution for LeetCode #687: Longest Univalue Path
/**
* 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 longestUnivaluePath(root: TreeNode | null): number {
let ans: number = 0;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
let [l, r] = [dfs(root.left), dfs(root.right)];
l = root.left && root.left.val === root.val ? l + 1 : 0;
r = root.right && root.right.val === root.val ? r + 1 : 0;
ans = Math.max(ans, l + r);
return 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.