Leetcode solutions: path sum I & II

Path sumgiven a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.


For example:
Given the below binary tree and sum = 22,
5
/\
4 8
//\
11 13 4
/\\
7 2 1
Return true, as there exist a root-to-leaf Path 5-> 4-> 11-> 2 which sum is 22.

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if(root==null)return false;if(root.left==null&&root.right==null)return sum==root.val?true:false;return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);    }}

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


For 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]
]

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {        ArrayList<ArrayList<Integer>> results=new ArrayList<ArrayList<Integer>>();if(root==null)return results;helper(results,new ArrayList<Integer>(),root,sum);return results;    }    private void helper(ArrayList<ArrayList<Integer>> results,ArrayList<Integer> path,TreeNode root,int sum){if(root==null)return;path.add(root.val);if(root.left==null&&root.right==null){if(root.val==sum){ArrayList<Integer> temp=new ArrayList<Integer>(path);results.add(temp);}}helper(results,path,root.left,sum-root.val);helper(results,path,root.right,sum-root.val);path.remove(path.size()-1);}}

Leetcode solutions: path sum I & II