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
转载于:https://www.cnblogs.com/vercont/p/10210302.html
