path.remove, Binary Tree Path Sum

Question

Have a question about a basic java sample for binary tree: Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.(A valid path is from root node to any of the leaf nodes.). Why we need to path.remove(path.size() - 1);when it go through the left, right node?

Here is the code sample:

public class Solution {

    public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {


        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }

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

    private void helper(TreeNode root,
                        ArrayList<Integer> path,
                        int sum,
                        int target,
                        List<List<Integer>> result) {

        // meet leaf
        if (root.left == null && root.right == null) {
            if (sum == target) {
                result.add(new ArrayList<Integer>(path));
            }
            return;
        }

        // go left
        if (root.left != null) {
            path.add(root.left.val);
            helper(root.left, path, sum + root.left.val, target, result);
            path.remove(path.size() - 1);
        }

        // go right
        if (root.right != null) {
            path.add(root.right.val);
            helper(root.right, path, sum + root.right.val, target, result);
            path.remove(path.size() - 1);
        }
    }
}

象嘉道象嘉道 · Accepted Answer · 2016-05-11T17:48:12

in this case, path is a temporary facility. after visits the left and starts processing the right, if the prog doesn't clean root.right.val from path, it goes incorrect.

path.remove, Binary Tree Path Sum

1 Answers