113. Path Sum II

Problem

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \    / \
7    2  5   1

Return:

[
   [5,4,11,2],
   [5,8,4,5]
]

Solution (Java)

/**
 * 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 {
    public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        recur(res, new ArrayList<>(), 0, targetSum, root);
        return res;
    }

    private void recur(
            List<List<Integer>> res, ArrayList<Integer> al, int sum, int targetSum, TreeNode root) {
        if (root == null) {
            return;
        }
        al.add(root.val);
        sum += root.val;
        if (sum == targetSum && root.left == null && root.right == null) {
            res.add(new ArrayList<>(al));
        }
        recur(res, al, sum, targetSum, root.left);
        recur(res, al, sum, targetSum, root.right);
        al.remove(al.size() - 1);
    }
}

Solution (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} sum
 * @return {number[][]}
 */
var pathSum = function(root, sum) {
  var res = [];
  helper(root, sum, [], res);
  return res;
};

var helper = function (root, sum, now, res) {
  if (!root) return;

  now.push(root.val);

  if (root.val === sum && !root.left && !root.right) res.push(now);

  helper(root.left, sum - root.val, Array.from(now), res);
  helper(root.right, sum - root.val, Array.from(now), res);
};

Explain:

nope.

Complexity:

• Time complexity :
• Space complexity :