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 and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.
The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
Example 1:
Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8 Output: 3 Explanation: The paths that sum to 8 are shown.
Example 2:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 Output: 3
Constraints:
[0, 1000].-109 <= Node.val <= 109-1000 <= targetSum <= 1000Problem summary: Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum. The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[10,5,-3,3,2,null,11,3,-2,null,1] 8
[5,4,8,11,null,13,4,7,2,null,null,5,1] 22
path-sum)path-sum-ii)path-sum-iv)longest-univalue-path)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #437: Path Sum III
/**
* 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 Map<Long, Integer> cnt = new HashMap<>();
private int targetSum;
public int pathSum(TreeNode root, int targetSum) {
cnt.put(0L, 1);
this.targetSum = targetSum;
return dfs(root, 0);
}
private int dfs(TreeNode node, long s) {
if (node == null) {
return 0;
}
s += node.val;
int ans = cnt.getOrDefault(s - targetSum, 0);
cnt.merge(s, 1, Integer::sum);
ans += dfs(node.left, s);
ans += dfs(node.right, s);
cnt.merge(s, -1, Integer::sum);
return ans;
}
}
// Accepted solution for LeetCode #437: Path Sum III
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func pathSum(root *TreeNode, targetSum int) int {
cnt := map[int]int{0: 1}
var dfs func(*TreeNode, int) int
dfs = func(node *TreeNode, s int) int {
if node == nil {
return 0
}
s += node.Val
ans := cnt[s-targetSum]
cnt[s]++
ans += dfs(node.Left, s) + dfs(node.Right, s)
cnt[s]--
return ans
}
return dfs(root, 0)
}
# Accepted solution for LeetCode #437: Path Sum III
# 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 pathSum(self, root: Optional[TreeNode], targetSum: int) -> int:
def dfs(node, s):
if node is None:
return 0
s += node.val
ans = cnt[s - targetSum]
cnt[s] += 1
ans += dfs(node.left, s)
ans += dfs(node.right, s)
cnt[s] -= 1
return ans
cnt = Counter({0: 1})
return dfs(root, 0)
// Accepted solution for LeetCode #437: Path Sum III
// 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::collections::HashMap;
use std::rc::Rc;
impl Solution {
pub fn path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> i32 {
let mut cnt = HashMap::new();
cnt.insert(0, 1);
fn dfs(
node: Option<Rc<RefCell<TreeNode>>>,
s: i64,
target: i64,
cnt: &mut HashMap<i64, i32>,
) -> i32 {
if let Some(n) = node {
let n = n.borrow();
let s = s + n.val as i64;
let ans = cnt.get(&(s - target)).copied().unwrap_or(0);
*cnt.entry(s).or_insert(0) += 1;
let ans = ans
+ dfs(n.left.clone(), s, target, cnt)
+ dfs(n.right.clone(), s, target, cnt);
*cnt.get_mut(&s).unwrap() -= 1;
ans
} else {
0
}
}
dfs(root, 0, target_sum as i64, &mut cnt)
}
}
// Accepted solution for LeetCode #437: Path Sum III
/**
* 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 pathSum(root: TreeNode | null, targetSum: number): number {
const cnt: Map<number, number> = new Map();
const dfs = (node: TreeNode | null, s: number): number => {
if (!node) {
return 0;
}
s += node.val;
let ans = cnt.get(s - targetSum) ?? 0;
cnt.set(s, (cnt.get(s) ?? 0) + 1);
ans += dfs(node.left, s);
ans += dfs(node.right, s);
cnt.set(s, (cnt.get(s) ?? 0) - 1);
return ans;
};
cnt.set(0, 1);
return dfs(root, 0);
}
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.