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]
]

这题可用全子集的递归模板（即完成后再从后面减掉），还有一个重点则是path是从root 到 leave， leave的判断条件是左右子树为Null

还有就是往最后结果集合中加元素时，一定要hard copy

public class Solution {
    /**
     * @param root the root of binary tree
     * @param target an integer
     * @return all valid paths
     */
    private int sum = 0;
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        // Write your code here
        List<List<Integer>> ans = new ArrayList<>();
        List<Integer> cur = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        sum = root.val;
        cur.add(root.val);
        bTreePathSumHelper(ans, cur, root, target);
        return ans;
    }

    public void bTreePathSumHelper(List<List<Integer>> ans, 
                                   List<Integer> cur, 
                                   TreeNode root,
                                   int target) {
        if (root.left == null && root.right == null) {
            if (sum == target) {
                ans.add(new ArrayList<Integer>(cur));
            }
            return;
        }

        if (root.left != null) {
            sum = sum + root.left.val;
            cur.add(root.left.val);
            bTreePathSumHelper(ans, cur, root.left, target);
            sum = sum - cur.get(cur.size() - 1);
            cur.remove(cur.size() - 1);
        }

        if (root.right != null) {
            sum = sum + root.right.val;
            cur.add(root.right.val);
            bTreePathSumHelper(ans, cur, root.right, target);
            sum = sum - cur.get(cur.size() - 1);
            cur.remove(cur.size() - 1);
        }

    }
}

second batch 判断时一定要加上判断当前节点是否是根节点（root.left == null && root.right == null)

int sum = 0;
    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
        // Write your code here
        List<List<Integer>> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }

        List<Integer> cur = new ArrayList<>();
        btpHelper(ans, cur, root, target);
        return ans;


    }

    public void btpHelper(List<List<Integer>> ans, List<Integer> cur, TreeNode root, int target) {
        if (root == null) {
            return;
        }

        cur.add(root.val);
        sum += root.val;

        btpHelper(ans, cur, root.left, target);
        btpHelper(ans, cur, root.right, target);

        if (root.left == null && root.right == null && sum == target)  {
            ans.add(new ArrayList<>(cur));
        }

        sum -= cur.get(cur.size() - 1);
        cur.remove(cur.size() - 1);

    }