Binary Tree Path Sum Lintcode

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.

Have you met this question in a real interview? 

Yes

Example

Given a binary tree, and target = 5:

     1
    / \
   2   4
  / \
 2   3

return

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

这道题居然一遍bug free了。。。但好像空间复杂度有点高。。。

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) {
        List<List<Integer>> res = new ArrayList<>();
        
        helper(root, target, res, new ArrayList<Integer>());
        return res;
    }
    public TreeNode helper(TreeNode root, int target, List<List<Integer>> res, List<Integer> sub) {
        if (root == null) {
            return null;
        }
        sub.add(root.val);
        
        TreeNode left = helper(root.left, target - root.val, res, new ArrayList<>(sub));
        TreeNode right = helper(root.right, target - root.val, res, new ArrayList<>(sub));
        
        if (left == null && right == null && target - root.val == 0) {
            res.add(sub);
        }
        
        return root;
        
    }
}

看了下答案，原来还可以改进一下，每次回去的时候把添加的值去掉，这样就不用不停地建新的了。对递归的认识好像又深了一点呢。

 

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) {
        List<List<Integer>> res = new ArrayList<>();
        
        helper(root, target, res, new ArrayList<Integer>());
        return res;
    }
    public void helper(TreeNode root, int target, List<List<Integer>> res, List<Integer> sub) {
        if (root == null) {
            return;
        }
        sub.add(root.val);
        
        if (root.left == null && root.right == null && target - root.val == 0) {
            res.add(new ArrayList<>(sub));
            return;
        }
        helper(root.left, target - root.val, res, sub);
        if (root.left != null) {
            sub.remove(sub.size() - 1);
        }
        helper(root.right, target - root.val, res, sub);
        if (root.right != null) {
            sub.remove(sub.size() - 1);
        }

    }
}

今天又做了一遍这道题，虽然思路还是会的但是居然写最优解还是卡住了！心酸啊，要注意只要递归的不是null就说明这个点被加过了所以要remove掉。如果是null的话就没有被加过，就不要平白无故的remove了。

另外这道题居然也可以用linkedlist和stack来解，也可以返回linkedlist。思路差不多，在此不表。

还是回顾一下吧。。。心酸。。。