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[LeetCode] 113. 路径总和 II

题目链接 : https://leetcode-cn.com/problems/path-sum-ii/

题目描述:

给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。

说明: 叶子节点是指没有子节点的节点。

示例:

给定如下二叉树，以及目标和 sum = 22，

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

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

思路:

和上一题一样, 用DFS只不过在遍历时候,要记录val而已

def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
        res = []
        if not root: return []
        def helper(root,sum, tmp):
            if not root:
                return 
            if not root.left and not root.right and sum - root.val == 0 :
                tmp += [root.val]
                res.append(tmp)
                return 
            helper(root.left, sum - root.val, tmp + [root.val])
            helper(root.right, sum - root.val, tmp + [root.val])
        helper(root, sum, [])
        return res

java

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<>();
        helper(root, sum, res, new ArrayList<Integer>());
        return res;
    }

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