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# 113.Path-Sum-II

## 题目描述

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]
]

## 代码

### Approach #1 Backtracking

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> ans = new ArrayList<List<Integer>>();
List<Integer> pathNodes = new ArrayList<Integer>();
dfs(root, sum, pathNodes, ans);
return ans;
}
private void dfs(TreeNode node, int remainingSum, List<Integer> pathNodes, List<List<Integer>> pathsList) {
if (node == null) return;
// Check if the current node is a leaf and also, if it
// equals our remaining sum. If it does, we add the path to
// our list of paths
if (remainingSum == node.val && node.left == null && node.right == null) {