Lintcode376-Binary Tree Path Sum-Easy

Posted 2021-02-15 jessiezyr

376. Binary Tree Path Sum

中文English

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.

Example

Example 1:

Input:
{1,2,4,2,3}
5
Output: [[1, 2, 2],[1, 4]]
Explanation:
The tree is look like this:
	     1
	    / 	   2   4
	  / 	 2   3
For sum = 5 , it is obviously 1 + 2 + 2 = 1 + 4 = 5

Example 2:

Input:
{1,2,4,2,3}
3
Output: []
Explanation:
The tree is look like this:
	     1
	    / 	   2   4
	  / 	 2   3
Notice we need to find all paths from root node to leaf nodes.
1 + 2 + 2 = 5, 1 + 2 + 3 = 6, 1 + 4 = 5 
There is no one satisfying it.

思路：

递归法，用helper方法，可以传更多参数

注意：

代码：

/*
     * @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>> result = new ArrayList<>();
        ArrayList<Integer> path = new ArrayList<>();
        
        if (root == null) {
            return result;
        }
        path.add(root.val);
        helper(root, path, root.val, target, result);
        return result;
    }
    
    public void helper (TreeNode root,
                        ArrayList<Integer> path,
                        int sum,
                        int target,
                        List<List<Integer>> result) {
        //meet leaf
        if (root.right == null && root.left == null) {
            if (sum == target) {
                result.add(new ArrayList(path));
            }
            return;
        }
        // right node
        if (root.right != null) {
            path.add(root.right.val);
            helper(root.right, path, sum + root.right.val, target, result);
            path.remove(path.size() - 1);
        }
        //left node
        if (root.left != null) {
            path.add(root.left.val);
            helper(root.left, path, sum + root.left.val, target, result);
            path.remove(path.size() - 1);
        }
    }