376. Binary Tree Path Sum [LintCode]

Given a binary tree, find all paths that sum of the nodes in the path equals to a given numbertarget.

A valid path is from root node to any of the leaf nodes.

Example

Given a binary tree, and target =5:

     1
    / \
   2   4
  / \
 2   3

return

[
  [1, 2, 2],
  [1, 4]
]

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root the root of binary tree
     * @param target an integer
     * @return all valid paths
     */
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        // Write your code here

        List<List<Integer>> results = new ArrayList();
        if(root == null) return results;

        List<Integer> path = new ArrayList();
        path.add(root.val);
        helper(root, root.val, target, path, results);
        return results;
    }

    private void helper(TreeNode node, int sum, int target, List<Integer> path, List<List<Integer>> results){

        if(node.left == null && node.right == null){

            if(sum == target){
               results.add(new ArrayList<Integer>(path));
            }
            return;
        }

        if(node.left != null){

            path.add(node.left.val);
            helper(node.left, sum + node.left.val, target, path, results);
            path.remove(path.size() - 1);
        }

        if(node.right != null){

            path.add(node.right.val);
            helper(node.right, sum + node.right.val, target, path, results);
            path.remove(path.size() - 1);
        }
    }
}