LEETCODE —— Binary Tree的3 题 —— 3种非Recursive遍历
Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1
\
2
/
3
return [1,2,3]
.
Note: Recursive solution is trivial, could you do it iteratively?
'''
Created on Nov 18, 2014 @author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com>
'''
# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None class Solution:
# @param root, a tree node
# @return a list of integers
def preorderTraversal(self, root):
stack=[]
vals=[]
if(root==None): return vals
node=root
stack.append(node)
while(len(stack)!=0):
node=stack.pop()
if(node==None): continue
vals.append(node.val)
stack.append(node.right)
stack.append(node.left) return vals
Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1
\
2
/
3
return [1,3,2]
.
Note: Recursive solution is trivial, could you do it iteratively?
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
'''
Created on Nov 18, 2014 @author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com>
'''
# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None class Solution:
# @param root, a tree node
# @return a list of integers
def inorderTraversal(self, root):
stack=[]
vals=[]
visited={}
if(root==None): return vals
node=root
stack.append(node)
visited[node]=1
while(len(stack)!=0):
if(node.left!=None and visited.has_key(node.left)==False):
node=node.left
stack.append(node)
visited[node]=1
else:
node=stack.pop()
if(node==None): continue
vals.append(node.val)
if(node.right!=None):
stack.append(node.right)
node=node.right
return vals
Binary Tree Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1
\
2
/
3
return [3,2,1]
.
Note: Recursive solution is trivial, could you do it iteratively?
'''
Created on Nov 19, 2014 @author: ScottGu<gu.kai.66@gmail.com, 150316990@qq.com>
'''
# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None class Solution:
# @param root, a tree node
# @return a list of integers
def postorderTraversal(self, root):
visited={}
stack=[]
vals=[]
if(root==None): return vals
node=root stack.append(node)
visited[node]=1 while(len(stack)!=0):
node=stack[-1]
if(node.left !=None and visited.has_key(node.left)==False):
stack.append(node.left)
visited[node.left]=1
continue
else:
if(node.right!=None and visited.has_key(node.right)==False):
stack.append(node.right)
visited[node.right]=1
continue
node=stack.pop()
if(node==None): continue
vals.append(node.val) return vals
LEETCODE —— Binary Tree的3 题 —— 3种非Recursive遍历的更多相关文章
- [LeetCode] Binary Tree Vertical Order Traversal 二叉树的竖直遍历
Given a binary tree, return the vertical order traversal of its nodes' values. (ie, from top to bott ...
- Binary Tree的3种非Recursive遍历
Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' valu ...
- 145.Binary Tree Postorder Traversal---二叉树后序非递归遍历
题目链接 题目大意:后序遍历二叉树. 法一:普通递归,只是这里需要传入一个list来存储遍历结果.代码如下(耗时1ms): public List<Integer> postorderTr ...
- 94.Binary Tree Inorder Traversal---二叉树中序非递归遍历
题目链接 题目大意:中序遍历二叉树.先序见144,后序见145. 法一:DFS,没啥说的,就是模板DFS.代码如下(耗时1ms): public List<Integer> inorder ...
- LeetCode:Binary Tree Level Order Traversal I II
LeetCode:Binary Tree Level Order Traversal Given a binary tree, return the level order traversal of ...
- LeetCode: Binary Tree Traversal
LeetCode: Binary Tree Traversal 题目:树的先序和后序. 后序地址:https://oj.leetcode.com/problems/binary-tree-postor ...
- LeetCode Binary Tree Paths(简单题)
题意: 给出一个二叉树,输出根到所有叶子节点的路径. 思路: 直接DFS一次,只需要判断是否到达了叶子,是就收集答案. /** * Definition for a binary tree node. ...
- [LeetCode] Binary Tree Level Order Traversal 与 Binary Tree Zigzag Level Order Traversal,两种按层次遍历树的方式,分别两个队列,两个栈实现
Binary Tree Level Order Traversal Given a binary tree, return the level order traversal of its nodes ...
- [LeetCode] Binary Tree Longest Consecutive Sequence 二叉树最长连续序列
Given a binary tree, find the length of the longest consecutive sequence path. The path refers to an ...
随机推荐
- iOS 8 TabBar 图片显示真实颜色
“展信颜开” 我怎么想到这个词了呢……因为这个足以表达我现在的心情,有解决了一个问题,有了一个收获. 早上小伙伴问我“用自带的tab改图的颜色他会不显示?改tabitem.”我记得是可以显示的,但是他 ...
- Java泛型学习笔记 - (七)浅析泛型中通配符的使用
一.基本概念:在学习Java泛型的过程中, 通配符是较难理解的一部分. 主要有以下三类:1. 无边界的通配符(Unbounded Wildcards), 就是<?>, 比如List< ...
- 将word文件快速转换成表格的技巧
最近烦心事还真是很多,世界买家网最近就遇到了很多烦心事. 从www.buyerinfo.biz网站中的数据导出为csv格式的文件,我导出了buyer的数据,那怎么把它制作成表格呢? 找了下,发现还是比 ...
- 近期编程问题——epoll failed:bad file descriptor
出现问题:epoll_wait:Bad file descriptor 原因:IO时间的socket描述符在epoll_ctl处理前就关闭了. 解决方法:不要在epoll_ctl之前关闭socket描 ...
- spring aop 声明式事务管理
一.声明式事务管理的概括 声明式事务(declarative transaction management)是Spring提供的对程序事务管理的方式之一. Spring的声明式事务顾名思义就是采用声明 ...
- swift 构建类
参开 http://blog.csdn.net/chelongfei/article/details/49784633 在 Swift 中, 类的初始化有两种方式, 分别是 Designated In ...
- Noi2011 阿狸的打字机
..] of longint; e,q,fa,ps,pt,fail,ans:..] of longint; trie:..,..] of longint; c:..] of longint; s:.. ...
- sass/less/stylus css编译
早上来了听一同事说stylus如何才能编译成css文件,瞬时间有点蒙,一听感觉和less是差不多的功能,随着就上网去查,然后发现这个文章,介绍了这三种sass/less/stylus的安装和语法,贴在 ...
- Ubuntu 16.04 64位安装insight 6.8
1. apt-get install insight已经不管用. 2. 编译源码死都有问题. 3. 拜拜,用KDBG.
- spring.net 如何让xml智能提示
tools ->options ->Text Editor ->xml ->miscellaneous ->network ->勾选 automatically d ...