Leetcode-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]
]

Solution:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<Integer> curPath = new ArrayList<Integer>();
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (root==null) 
            return res;
        pathSumRecur(root,sum,curPath,res);
        return res;
    }
    
    public void pathSumRecur(TreeNode curNode, int residual, List<Integer> curPath, List<List<Integer>> res){
        if (curNode.left==null&&curNode.right==null){
            if (residual==curNode.val){
                List<Integer> newPath = new ArrayList<Integer>();
                newPath.addAll(curPath);
                newPath.add(curNode.val);
                res.add(newPath);
                return; 
            } else 
               return;
        }
       
        curPath.add(curNode.val);
        if (curNode.left!=null)
            pathSumRecur(curNode.left,residual-curNode.val,curPath,res);
        if (curNode.right!=null)
            pathSumRecur(curNode.right,residual-curNode.val,curPath,res);
        curPath.remove(curPath.size()-1);
    }
}

This is a recursive problem. Record the path from root to current node. If find a valid path at some leaf node, then create a copy of current path, and put it into the result set.

//NOTE: since the curPath is changed at each level, we need to restore it after running the recursion function.