Binary Tree stored in acmclub 1757 binary linked list-from lanshui_yang

Description:

In this article, we will provide a string that is traversed in the First Order. spaces represent empty subnodes, and uppercase letters represent node content. Create a binary tree using this string and output each non-empty node according to one of the first traversal and two middle Traversal Algorithms described in the topic.

The input format contains only one row, which contains a string S, used to create a binary tree. Ensure that S is a valid binary tree that traverses strings in sequence. The node content only has uppercase letters, and the length of S cannot exceed 100. The output consists of three lines. Each line contains a string of characters, indicating the node content obtained in the first, middle, and middle order respectively. Each letter is followed by a space. Note the line feed at the end of the line. Sample Input

ABC de G F

Sample output

A B C D E G F
C B e g d F
C B e g d F

Solution: there is a difficulty and a pitfall in this question! The difficulty is the problem of input and build. For specific solutions, see the program. There are two ways to solve the problem: direct recursion (relatively simple, not detailed); Type 2: manually call the stack (see the program if you need to pay attention to it !!)

The Program for direct recursive traversal is as follows:

# Include <cstdio> # include <string> # include <algorithm> # include <cstring> # include <cstdlib> # include <iostream> using namespace STD; const int maxn = 1111111; struct node {char data; int Xu; node * left; node * right; node (): Data ('\ 0'), Xu (0), left (null ), right (null) {};}; int flag; node * STAP [maxn]; int top; void creattree (node * & T) // note: input and build methods here {If (FLAG) return; char TMP; TMP = getchar (); If (TMP = '') {T = NULL ;} else if (TMP = '\ n') {flag = 1; return;} else {T = new node; t-> DATA = TMP; creattree (t-> left); creattree (t-> right) ;}} void initv (node * P) {If (P) {cout <p-> data <"; initv (p-> left); initv (p-> right) ;}} void midv (node * P) {If (p) {midv (p-> left); cout <p-> data <""; midv (p-> right) ;}} int main () {flag = 0; node * head; creattree (head); initv (head); cout <Endl; midv (head ); cout <Endl; return 0 ;}

Method 2: manually call the stack

# Include <cstdio> # include <string> # include <algorithm> # include <cstring> # include <cstdlib> # include <iostream> using namespace STD; const int maxn = 1111111; struct node {char data; int Xu; node * left; node * right; node (): Data ('\ 0'), Xu (0), left (null ), right (null) {};}; int flag; node * STAP [maxn]; int top; void creattree (node * & T) {If (FLAG) return; char TMP; TMP = getchar (); If (TMP = '') {T = NULL;} else if (TMP = '\ n') {flag = 1; return;} else {T = new node; t-> DATA = TMP; creattree (t-> left); creattree (t-> right) ;}} void initv (node * P) {If (P) {cout <p-> data <"; initv (p-> left); initv (p-> right) ;}} void mid2v (node * P) {Top =-1; if (p) STAP [++ top] = P; node * TMP; while (top> = 0) {TMP = STAP [Top]; while (TMP) {STAP [++ top] = TMP-> left; TMP = TMP -> Left;} top --; // unstack the NULL pointer. // Note: This is very important, the empty pointer to stack does not mean that the Left subtree of the node shown in the above program has been accessed. // It may also be the right subtree of the node shown below !! If (top> = 0) {TMP = STAP [top --]; cout <TMP-> data <""; STAP [++ top] = TMP-> right ;}} int main () {flag = 0; node * head; creattree (head); initv (head ); cout <Endl; mid2v (head); cout <Endl; mid2v (head); cout <Endl; return 0 ;}