113 Path Sum II.java

// 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

import java.util.ArrayList;
import java.util.List;

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;
        
        dfs(root, targetSum, new ArrayList<>(), res);
        return res;
    }
    private static void dfs(TreeNode root, int target, List<Integer> list, List<List<Integer>> res){
        if(root == null)    
            return;

        list.add(root.val);
        
        if(root.left == null && root.right == null && target == root.val ){
            res.add(list);
            return;
        }
        
        dfs(root.left, target - root.val, new ArrayList<>(list), res);
        dfs(root.right, target - root.val, new ArrayList<>(list), res);
    }
}