leetcode第一天-merge two binary trees
有段时间没有写代码了,脑子都生锈了,今后争取笔耕不辍(立flag,以后打脸)
随机一道Leecode题, Merge Two Binary Trees,题目基本描述如下:
Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.
You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.
题目是很简单的题目,但是由于大脑生锈,竟然也折腾了半天。
解题思路是使用迭代。
令tree1为t1,tree2为t2
merge tree的root肯定为t1.root+t2.root
对于merge tree的left,可以认为是以t1.left为root的tree与以t2.left为root的tree合并的结果
同样类似,对于merge tree的right,可以认为是t1.right为root的tree与以t2.right为root的tree合并的结果
如此反复,通过迭代就很容易得到结果了。
做个备忘,在python3中,使用TreeNode来构建树,树当前的节点的值为val,节点的左子节点为left,节点的右子为right.
代码如下:
class Solution:
def mergeTrees(self, t1, t2):
"""
:type t1: TreeNode
:type t2: TreeNode
:rtype: TreeNode
"""
if (t1 is None) and (t2 is None):
return
if (t1 is None):
return t2
if (t2 is None):
return t1
t1.val += t2.val
t1.left = self.mergeTrees(t1.left,t2.left)
t1.left = self.mergeTrees(t1.right,t2.right)
return t1
def stringToTreeNode(input):
input = input.strip()
input = input[1:-1]
if not input:
return None
inputValues = [s.strip() for s in input.split(',')]
root = TreeNode(int(inputValues[0]))
nodeQueue = [root]
front = 0
index = 1
while index < len(inputValues):
node = nodeQueue[front]
front = front + 1
item = inputValues[index]
index = index + 1
if item != "null":
leftNumber = int(item)
node.left = TreeNode(leftNumber)
nodeQueue.append(node.left)
if index >= len(inputValues):
break
item = inputValues[index]
index = index + 1
if item != "null":
rightNumber = int(item)
node.right = TreeNode(rightNumber)
nodeQueue.append(node.right)
return root
def treeNodeToString(root):
if not root:
return "[]"
output = ""
queue = [root]
current = 0
while current != len(queue):
node = queue[current]
current = current + 1
if not node:
output += "null, "
continue
output += str(node.val) + ", "
queue.append(node.left)
queue.append(node.right)
return "[" + output[:-2] + "]"
def main():
import sys
def readlines():
for line in sys.stdin:
yield line.strip('\n')
lines = readlines()
while True:
try:
line = next(lines)
t1 = stringToTreeNode(line);
line = next(lines)
t2 = stringToTreeNode(line);
ret = Solution().mergeTrees(t1, t2)
out = treeNodeToString(ret);
print(out)
except StopIteration:
break
if __name__ == '__main__':
main()
leetcode第一天-merge two binary trees的更多相关文章
- 【LeetCode】617. Merge Two Binary Trees 解题报告
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 递归 日期 题目地址:https://leetcod ...
- 【leetcode】617. Merge Two Binary Trees
原题 Given two binary trees and imagine that when you put one of them to cover the other, some nodes o ...
- LeetCode算法题-Merge Two Binary Trees(Java实现)
这是悦乐书的第274次更新,第290篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第142题(顺位题号是617).提供两个二叉树,将其合并为新的二叉树,也可以在其中一个二 ...
- LeetCode 617. 合并二叉树(Merge Two Binary Trees)
617. 合并二叉树 617. Merge Two Binary Trees 题目描述 给定两个二叉树,想象当你将它们中的一个覆盖到另一个上时,两个二叉树的一些节点便会重叠. 你需要将他们合并为一个新 ...
- 【Leetcode_easy】617. Merge Two Binary Trees
problem 617. Merge Two Binary Trees 参考 1. Leetcode_easy_617. Merge Two Binary Trees; 完
- Week2 - 669. Trim a Binary Search Tree & 617. Merge Two Binary Trees
Week2 - 669. Trim a Binary Search Tree & 617. Merge Two Binary Trees 669.Trim a Binary Search Tr ...
- [LeetCode] Merge Two Binary Trees 合并二叉树
Given two binary trees and imagine that when you put one of them to cover the other, some nodes of t ...
- [LeetCode] 617. Merge Two Binary Trees 合并二叉树
Given two binary trees and imagine that when you put one of them to cover the other, some nodes of t ...
- LeetCode 617 Merge Two Binary Trees 解题报告
题目要求 Given two binary trees and imagine that when you put one of them to cover the other, some nodes ...
随机推荐
- Hillstone设备管理-恢复出厂设置
1.CLI命令行操作 unset all: 根据提示选择是否保存当前配置y/n: 选择是否重启y/n: 系统重启后即恢复到出厂设置. 2.webUI操作 “系统”—“配置”,点击“清除”按钮,系统会提 ...
- mysql学习3:mysql之my.cnf详解
mysql之my.cnf详解 本文转自:https://www.cnblogs.com/panwenbin-logs/p/8360703.html 以下是 my.cnf 配置文件参数解释: #*** ...
- 不常用但是很实用的css记录
本文主旨是记录一些不常用但是非常炫酷的css属性,提升用户体验的捷径之一. 1.background-attachment 滚动视差 https://codepen.io/Chokcoco/p ...
- H5真机调试
为什么要做H5真机调试? 第一,样式调试.浏览器的效果和真机上的效果不一定相同,没有真机调试,我们都是先上传到服务器,然后再用手机访问,显然对于样式调试来说,这样效率很低. 第二,调用到底层.真机才拥 ...
- Swoole addProcess的使用
addProcess函数 是添加一个用户自定义的工作进程.这个有什么用呢?服务在启动后,可以用于监控.上报或者其他特殊的任务. 注意这个添加的进程是被manager进程管理的.如果这个添加的用户进程经 ...
- rabbit基本原理 转
https://www.cnblogs.com/jun-ma/p/4840869.html
- memcache集群
实现memcache集群 一:memcache本身没有redis锁具备的数据持久化功能,比如RDB和AOF都没有,但是可以通过做集群的方式,让各memcache的数据进行同步,实现数据的一致性,即 ...
- mysql sql mode
/usr/local/mysql/bin/mysqld --verbose --help | grep -A 1 'Default options' (1)关于配置文件路径 有时候,我发现虽然尝试修改 ...
- C语言错题集
1.输入两个int型数a.b,求a/b的商c,不必考虑b为0的情况,输出c(保留两位小数) include<stdio.h> int main() { int a,b; float c; ...
- 关于TypeError: strptime() argument 1 must be str, not bytes解析
关于TypeError: strptime() argument 1 must be str, not bytes解析 在使用datetime.strptime(s,fmt)来输出结果日期结果时, ...