Path Sum II

Problem

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

Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {number[][]}
 */
var pathSum = function(root, targetSum) {
    const res = [];
    const dfs = (node, target, path) => {
        if (!node) {
            return;
        }
        const sum = target - node.val;
        path.push(node.val);
        if (node.left || node.right) { // it not leaf node
            dfs(node.left, sum, path);
            dfs(node.right, sum, path);
        } else if (!sum) { // is leaf node
            res.push([...path]);
        }
        path.pop();
    }
    dfs(root, targetSum, []);
    return res;
};

We will implement a DFS solution. We traverse the tree root while keeping track of our path path. When we reach a leaf node, check if targetSum is 0, if so, add path to our resulting set res.

Problem​

Solution​

Problem

Solution