输出单层结点 牛客网 程序员面试金典 C++ Python

输出单层结点 牛客网 程序员面试金典 C++ Python

• 题目描述

• 对于一棵二叉树，请设计一个算法，创建含有某一深度上所有结点的链表。

• 给定二叉树的根结点指针TreeNode* root，以及链表上结点的深度，请返回一个链表ListNode，代表该深度上所有结点的值，请按树上从左往右的顺序链接，保证深度不超过树的高度，树上结点的值为非负整数且不超过100000。

C++

/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};*/
class TreeLevel {
public:
    //run:4ms memory:476k
    ListNode* head = new ListNode(-1);
    ListNode* p = head;
    ListNode* getTreeLevel(TreeNode* root, int dep) {
        if (NULL == root || dep <= 0) return NULL;
        if (1 == dep){
            ListNode* tmp = new ListNode(root->val);
            p->next = tmp;
            p = p->next;
        }else{
            getTreeLevel(root->left,dep - 1);
            getTreeLevel(root->right,dep - 1);
        }
        return head->next;
    }

Python

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class TreeLevel:
    #run:51ms memory:5856k
    head = ListNode(-1)
    p = head
    def getTreeLevel(self, root, dep):
        if None == root: return None
        if dep <= 0: return None
        if dep == 1:
            tmp = ListNode(root.val)
            self.p.next = tmp
            self.p = self.p.next
        else:
            self.getTreeLevel(root.left,dep-1)
            self.getTreeLevel(root.right,dep-1)
        return self.head.next