LeetCode #113 — MEDIUM

Path Sum II

Move from brute-force thinking to an efficient approach using backtracking strategy.

The Problem

Problem Statement

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.

A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: []

Example 3:

Input: root = [1,2], targetSum = 0
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 5000].
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000

Step 01

Brute Force Baseline

Problem summary: Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references. A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Backtracking · Tree

Example 1

[5,4,8,11,null,13,4,7,2,null,null,5,1]
22

Example 2

[1,2,3]
5

Example 3

[1,2]
0

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #113: Path Sum II
/**
 * 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 List<List<Integer>> ans = new ArrayList<>();
    private List<Integer> t = new ArrayList<>();

    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        dfs(root, targetSum);
        return ans;
    }

    private void dfs(TreeNode root, int s) {
        if (root == null) {
            return;
        }
        s -= root.val;
        t.add(root.val);
        if (root.left == null && root.right == null && s == 0) {
            ans.add(new ArrayList<>(t));
        }
        dfs(root.left, s);
        dfs(root.right, s);
        t.remove(t.size() - 1);
    }
}

// Accepted solution for LeetCode #113: Path Sum II
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) (ans [][]int) {
	t := []int{}
	var dfs func(*TreeNode, int)
	dfs = func(root *TreeNode, s int) {
		if root == nil {
			return
		}
		s -= root.Val
		t = append(t, root.Val)
		if root.Left == nil && root.Right == nil && s == 0 {
			ans = append(ans, slices.Clone(t))
		}
		dfs(root.Left, s)
		dfs(root.Right, s)
		t = t[:len(t)-1]
	}
	dfs(root, targetSum)
	return
}

# Accepted solution for LeetCode #113: Path Sum II
# 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) -> List[List[int]]:
        def dfs(root, s):
            if root is None:
                return
            s += root.val
            t.append(root.val)
            if root.left is None and root.right is None and s == targetSum:
                ans.append(t[:])
            dfs(root.left, s)
            dfs(root.right, s)
            t.pop()

        ans = []
        t = []
        dfs(root, 0)
        return ans

// Accepted solution for LeetCode #113: Path Sum II
// 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>>>,
        paths: &mut Vec<i32>,
        mut target_sum: i32,
        res: &mut Vec<Vec<i32>>,
    ) {
        if let Some(node) = root {
            let mut node = node.borrow_mut();
            target_sum -= node.val;
            paths.push(node.val);
            if node.left.is_none() && node.right.is_none() {
                if target_sum == 0 {
                    res.push(paths.clone());
                }
            } else {
                if node.left.is_some() {
                    Self::dfs(node.left.take(), paths, target_sum, res);
                }
                if node.right.is_some() {
                    Self::dfs(node.right.take(), paths, target_sum, res);
                }
            }
            paths.pop();
        }
    }

    pub fn path_sum(root: Option<Rc<RefCell<TreeNode>>>, target_sum: i32) -> Vec<Vec<i32>> {
        let mut res = vec![];
        let mut paths = vec![];
        Self::dfs(root, &mut paths, target_sum, &mut res);
        res
    }
}

// Accepted solution for LeetCode #113: Path Sum II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #113: Path Sum II
// /**
//  * 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 List<List<Integer>> ans = new ArrayList<>();
//     private List<Integer> t = new ArrayList<>();
// 
//     public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
//         dfs(root, targetSum);
//         return ans;
//     }
// 
//     private void dfs(TreeNode root, int s) {
//         if (root == null) {
//             return;
//         }
//         s -= root.val;
//         t.add(root.val);
//         if (root.left == null && root.right == null && s == 0) {
//             ans.add(new ArrayList<>(t));
//         }
//         dfs(root.left, s);
//         dfs(root.right, s);
//         t.remove(t.size() - 1);
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n!)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.

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.