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tree_level_node.py
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41 lines (36 loc) · 1.44 KB
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"""
链接：https://www.nowcoder.com/practice/cb6c883b123b44399377d0c71e6ba3ea?tpId=8&tqId=11014&rp=1&ru=%2Fta%2Fcracking-the-coding-interview&qru=%2Fta%2Fcracking-the-coding-interview%2Fquestion-ranking&tPage=2
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。
"""
# -*- coding:utf-8 -*-
# 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:
def getTreeLevel(self, root, dep):
queue=[(0,root)]
L_head=ListNode(0)
L_head_copy=ListNode(0)
def level(root,dep):
while queue:
lev,node=queue.pop(0)
if lev==dep:
Ln=ListNode(node.val)
L_head.next=Ln
L_head=Ln
if node.left == None:
queue.append((lev+1,node.left))
if node.right == None:
queue.append((lev+1,node.right))
return L_head_copy.next
# write code here