113 Path Sum II

113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example: Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

Company Tags: Bloomberg

分析&代码

和112 Path Sum一样，就是多了一个记录的功能

注意最后的一个叶节点的结果添加的时候，不要加到curList里面，如果加了，return之前要退掉

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null) {
            return res;
        }
        sumPath(root, sum, res, new ArrayList<Integer>());
        return res;
    }

    private void sumPath(TreeNode root, int target, List<List<Integer>> res, List<Integer> cur) {
        if(root == null) {
            return;
        }
        if(root.left == null && root.right == null) {
            if(root.val == target) {
                List<Integer> curRes = new ArrayList<>(cur);
                curRes.add(root.val);
                res.add(curRes);
            }
            return;
        }
        cur.add(root.val);
        sumPath(root.left, target - root.val, res, cur);
        sumPath(root.right, target - root.val, res, cur);
        cur.remove(cur.size() - 1);
    }