Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.
The path sum of a path is the sum of the node's values in the path.
Given the root of a binary tree, return the maximum path sum of any non-empty path.
Example 1:
Input: root = [1,2,3] Output: 6 Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.
Example 2:
Input: root = [-10,9,20,null,null,15,7] Output: 42 Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.
Constraints:
[1, 3 * 104].-1000 <= Node.val <= 1000Problem summary: A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root. The path sum of a path is the sum of the node's values in the path. Given the root of a binary tree, return the maximum path sum of any non-empty path.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming · Tree
[1,2,3]
[-10,9,20,null,null,15,7]
path-sum)sum-root-to-leaf-numbers)path-sum-iv)longest-univalue-path)time-needed-to-inform-all-employees)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #124: Binary Tree Maximum Path Sum
/**
* 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 = -1001;
public int maxPathSum(TreeNode root) {
dfs(root);
return ans;
}
private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int left = Math.max(0, dfs(root.left));
int right = Math.max(0, dfs(root.right));
ans = Math.max(ans, root.val + left + right);
return root.val + Math.max(left, right);
}
}
// Accepted solution for LeetCode #124: Binary Tree Maximum Path Sum
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func maxPathSum(root *TreeNode) int {
ans := -1001
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
left := max(0, dfs(root.Left))
right := max(0, dfs(root.Right))
ans = max(ans, left+right+root.Val)
return max(left, right) + root.Val
}
dfs(root)
return ans
}
# Accepted solution for LeetCode #124: Binary Tree Maximum Path Sum
# 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 maxPathSum(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode]) -> int:
if root is None:
return 0
left = max(0, dfs(root.left))
right = max(0, dfs(root.right))
nonlocal ans
ans = max(ans, root.val + left + right)
return root.val + max(left, right)
ans = -inf
dfs(root)
return ans
// Accepted solution for LeetCode #124: Binary Tree Maximum Path Sum
// 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>>>, res: &mut i32) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
let left = (0).max(Self::dfs(&node.left, res));
let right = (0).max(Self::dfs(&node.right, res));
*res = (node.val + left + right).max(*res);
node.val + left.max(right)
}
pub fn max_path_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut res = -1000;
Self::dfs(&root, &mut res);
res
}
}
// Accepted solution for LeetCode #124: Binary Tree Maximum Path Sum
/**
* 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 maxPathSum(root: TreeNode | null): number {
let ans = -1001;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
const left = Math.max(0, dfs(root.left));
const right = Math.max(0, dfs(root.right));
ans = Math.max(ans, left + right + root.val);
return Math.max(left, right) + root.val;
};
dfs(root);
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
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.