leetcode——113. 路径总和 II

同看懂做不出来。。。

class Solution:
    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

执行用时 :72 ms, 在所有 python3 提交中击败了39.94%的用户

内存消耗 :19.1 MB, 在所有 python3 提交中击败了5.08%的用户

 

——2019.11.21

 

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复习：

public List<List<Integer>> pathSum(TreeNode root, int sum) { //得用到栈吧
        //遍历所有路径
        List<List<Integer>> list = new ArrayList<>();
        if(root == null){
            return list;
        }
        Stack<Integer> path = new Stack<>();
        pathSum(root,sum,list,path);
        return list;
    }

    private void pathSum(TreeNode node, int sum, List<List<Integer>> list,Stack<Integer> path) {
        if(node == null && sum != 0){
            return;
        }
        if(node != null) {
            if (node.left == null && node.right == null && sum == node.val) {
                path.push(node.val);
                list.add(new ArrayList<>(path));
                path.pop();
            } else {
                path.push(node.val);
                pathSum(node.left, sum - node.val, list, path);
                pathSum(node.right, sum - node.val, list, path);
                path.pop();
            }
        }
    }

——2020.7.2