【遍历二叉树】11把二叉树转换成前序遍历的链表【Flatten Binary Tree to Linked List】
本质上是二叉树的root->right->left遍历。
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
给定一个二叉树,就地的把他转换成一个链表。
例如:
给定
1
/ \
2 5
/ \ \
3 4 6
转换后的树应该向这样子:
1
\
2
\
3
\
4
\
5
\
6
提示:
如果你观察的足够仔细,你会发现,每个还在的右孩子指针指向了,这个节点在前序遍历中的后面的那个节点。
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1
/ \
2 5
/ \ \
3 4 6
The flattened tree should look like:
1
\
2
\
3
\
4
\
5
\
6
If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order traversal.
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
迭代版本:
test.cpp:
|
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 |
#include <iostream>
#include <cstdio> #include <stack> #include <vector> #include "BinaryTree.h" using namespace std; /** TreeNode *tmp; if (tmp->right) tmp->left = NULL; // 树中结点含有分叉, ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); flatten(pNodeA1); TreeNode *trav = pNodeA1; DestroyTree(pNodeA1); |
递归版本:
test.cpp:
|
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 |
#include <iostream>
#include <cstdio> #include <stack> #include <vector> #include "BinaryTree.h" using namespace std; /** if(root == NULL) flatten(root->left); TreeNode *tmpright = root->right; return ; // 树中结点含有分叉, ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); flatten(pNodeA1); TreeNode *trav = pNodeA1; DestroyTree(pNodeA1); |
结果输出:
6 7 1 4 3 5 2
|
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_ struct TreeNode TreeNode *CreateBinaryTreeNode(int value); #endif /*_BINARY_TREE_H_*/ |
|
1
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 |
#include <iostream>
#include <cstdio> #include "BinaryTree.h" using namespace std; /** //创建结点 return pNode; //连接结点 //打印节点内容以及左右子结点内容 if(pNode->left != NULL) if(pNode->right != NULL) printf("\n"); //前序遍历递归方法打印结点内容 if(pRoot != NULL) if(pRoot->right != NULL) void DestroyTree(TreeNode *pRoot) delete pRoot; DestroyTree(pLeft); |
【遍历二叉树】11把二叉树转换成前序遍历的链表【Flatten Binary Tree to Linked List】的更多相关文章
- [Swift]LeetCode114. 二叉树展开为链表 | Flatten Binary Tree to Linked List
Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 ...
- leetcode 114. 二叉树展开为链表(Flatten Binary Tree to Linked List)
目录 题目描述: 示例: 解法: 题目描述: 给定一个二叉树,原地将它展开为链表. 示例: 给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4 \ 5 \ ...
- LeetCode 114| Flatten Binary Tree to Linked List(二叉树转化成链表)
题目 给定一个二叉树,原地将它展开为链表. 例如,给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4 \ 5 \ 6 解析 通过递归实现:可以用先序遍历, ...
- [LeetCode]Flatten Binary Tree to Linked List题解(二叉树)
Flatten Binary Tree to Linked List: Given a binary tree, flatten it to a linked list in-place. For e ...
- [LeetCode] Flatten Binary Tree to Linked List 将二叉树展开成链表
Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 T ...
- 114 Flatten Binary Tree to Linked List 二叉树转换链表
给定一个二叉树,使用原地算法将它 “压扁” 成链表.示例:给出: 1 / \ 2 5 / \ \ 3 4 6压扁后变成如下: ...
- [LeetCode] 114. Flatten Binary Tree to Linked List 将二叉树展开成链表
Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 T ...
- leetcode 114.Flatten Binary Tree to Linked List (将二叉树转换链表) 解题思路和方法
Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 ...
- [LeetCode] 114. Flatten Binary Tree to Linked List 将二叉树展平为链表
Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 ...
随机推荐
- 中国程序员如何去 Facebook 工作?
1.在Facebook,可以选择哪里工作? Facebook 在内地确实没有 Office ,但可以在https://www.facebook.com/careers/?ref=pf#location ...
- Apache禁止ip访问
网站突然让禁止ip访问,于是就通过配置Apache达到了想要的效果. 我们网站用的是Apache+tomcat集群,所以需要配置虚拟主机,虚拟主机我在这里就不说了,不明白的上网搜搜吧,这里只说禁止ip ...
- jQuery实现复选框全选/所有取消/反选/获得选择的值
<!DOCTYPE html> <html> <head> <script type="text/javascript" src=&quo ...
- 使用svn diff的-r参数的来比较任意两个版本的差异
1 svn diff的用法1.1 对比当前本地的工作拷贝文件(working copy)和缓存在.svn下的版本库文件的区别 svn diff 1.2 对比当前本地的工作拷贝文件(working co ...
- HTML 获取屏幕,浏览器,页面的高度
1,物理尺寸和分辨率 容器的尺寸是指当前分辨率下的高度.宽度,而不是物理高度.宽度. 如:一个22寸的显示器,屏幕分辨率为1366 * 768,那么获取到的屏幕高度为1366px,宽度为768px. ...
- 基于CocoaPods的iOS项目模块化实践
什么是CocoaPods? CocoaPods is a dependency manager for Swift and Objective-C Cocoa projects. It has ove ...
- python3 内置常用函数系列一
python3 内置了一系列的常用函数, python英文官方文档详细说明:点击查看, 为了方便查看,将内置常用的函数的记录一下来. Python3版本所有的内置函数: 1.abs() print(a ...
- 修改push动画的方向
CATransition *animation = [CATransition animation]; animation.duration = 0.4; animation.timingFuncti ...
- HDU OJ 2159 FATE
#include <stdio.h> #include <string.h> ][] ; ]; //»ñµÃ¾Ñé ]; //»¨·ÑµÄÈÌÄÍ¶È int main() ...
- docker网络模型
docker run -it --rm --net none --name test centos:newer /bin/bash --net none的作用是创建一个封闭的容器,容器只有lo接口,只 ...