文章目录
〇、前言
这两周开始跟着【MOOC-浙江大学-陈越、何钦铭-数据结构】进行数据结构与算法的学习，特此记录复习一下，虽然记不住，但是一直记一直记一直记，成为复读机就好了。
一、树的定义
二、树的基本术语
三、二叉树
四、性质
五、二叉树基本操作
六、二叉树遍历
1、先序遍历
2、中序遍历
3、后序遍历
4、层序遍历
七、树的同构
实现在后面的第一题！
八、课后题
1、03-树1 树的同构 (25分)
输入样例1（对应图1）：
8
A 1 2
B 3 4
C 5 -
D - -
E 6 -
G 7 -
F - -
H - -
8
G - 4
B 7 6
F - -
A 5 1
H - -
C 0 -
D - -
E 2 -
输出样例1:
Yes
输入样例2（对应图2）：
8
B 5 7
F - -
A 0 3
C 6 -
H - -
D - -
G 4 -
E 1 -
8
D 6 -
B 5 -
E - -
H - -
C 0 2
G - 3
F - -
A 1 4
输出样例2:
No
#include <stdio.h>
#define MaxTree 10
#define ElementType char
#define Tree int
#define Null -1
struct TreeNode{
ElementType Element;
Tree Left;
Tree Right;
}T1[MaxTree],T2[MaxTree];
Tree BuildTree(struct TreeNode T[]){
int N,i,check[MaxTree],Root=Null;
ElementType cl,cr;
scanf("%d\n",&N);
if(N){
for(i=0;i<N;i++)
check[i]=0;
for(i=0;i<N;i++){
scanf("%c %c %c\n",&T[i].Element,&cl,&cr);
if(cl!='-'){
T[i].Left=cl-'0';
check[T[i].Left]=1;
}
else{
T[i].Left=Null;
}
if(cr!='-'){
T[i].Right=cr-'0';
check[T[i].Right]=1;
}
else{
T[i].Right=Null;
}
}
for(i=0;i<N;i++){
if(!check[i]){
Root=i;
break;
}
}
}
return Root;
}
int Isomorphic(Tree R1,Tree R2){
//两树为空树，则为同构
if((R1==Null)&&(R2==Null))
return 1;
//一个树为空，另一个不为空，则不同构
if(((R1==Null)&&(R2!=Null))||((R1!=Null)&&(R2==Null)))
return 0;
//两树根结点存在但数据不同，则不同构
if(T1[R1].Element!=T2[R2].Element)
return 0;
//两树左子树均为空树，则判断两树右子树是否同构
if((T1[R1].Left==Null)&&(T2[R2].Left==Null))
return Isomorphic(T1[R1].Right,T2[R2].Right);
//两树左子树不为空，且两树左子树数据相等，则判断两树左子树和右子树是否同时同构
if((T1[T1[R1].Left].Element==T2[T2[R2].Left].Element)&&(T1[R1].Left!=Null)&&(T2[R2].Left!=Null))
return (Isomorphic(T1[R1].Left,T2[R2].Left)&&Isomorphic(T1[R1].Right,T2[R2].Right));
//否则交换判断子树是否同构
else
return (Isomorphic(T1[R1].Left,T2[R2].Right)&&Isomorphic(T1[R1].Right,T2[R2].Left));
}
int main(){
Tree R1,R2;
R1=BuildTree(T1);
R2=BuildTree(T2);
if(Isomorphic(R1,R2))
printf("Yes\n");
else
printf("No\n");
}
2、03-树2 List Leaves (25分)
Sample Input:
8
1 -
0 -
2 7
5 -
4 6
Sample Output:
4 1 5
#include <stdio.h>
#include <stdlib.h>
#define MaxTree 10
#define Tree int
#define Null -1
struct TreeNode {
Tree Left;
Tree Right;
} T1[MaxTree];
#define QueueSize 100
typedef int Position;
typedef int ElementType;
struct QNode {
ElementType *Data; /* 存储元素的数组 */
Position Front, Rear; /* 队列的头、尾指针 */
int MaxSize; /* 队列最大容量 */
};
typedef struct QNode *Queue;
Queue CreateQueue( int MaxSize )
{
Queue Q = (Queue)malloc(sizeof(struct QNode));
Q->Data = (ElementType *)malloc(MaxSize * sizeof(ElementType));
Q->Front = Q->Rear = 0;
Q->MaxSize = MaxSize;
return Q;
}
int IsFull( Queue Q )
{
return ((Q->Rear+1)%Q->MaxSize == Q->Front);
}
int AddQ( Queue Q, ElementType X )
{
if ( IsFull(Q) ) {
printf("队列满");
return;
}
else {
Q->Rear = (Q->Rear+1)%Q->MaxSize;
Q->Data[Q->Rear] = X;
return;
}
}
int IsEmpty( Queue Q )
{
return (Q->Front == Q->Rear);
}
ElementType DeleteQ( Queue Q )
{
if ( IsEmpty(Q) ) {
printf("队列空");
return Null;
}
else {
Q->Front =(Q->Front+1)%Q->MaxSize;
return Q->Data[Q->Front];
}
}
Tree BuildTree(struct TreeNode T[])
{
char cl, cr;
int N, i, check[MaxTree];
Tree Root = Null;
scanf("%d\n", &N);
if(N) {
for(i=0;i<N;i++)
check[i] = 0;
for(i=0;i<N;i++) {
scanf("%c %c\n", &cl, &cr);
if(cl!='-') {
T1[i].Left = cl - '0';
check[T1[i].Left] = 1;
} else
T1[i].Left = Null;
if(cr!='-') {
T1[i].Right = cr - '0';
check[T1[i].Right] = 1;
} else
T1[i].Right = Null;
}
for(i=0;i<N;i++)
if(check[i] != 1) break;
Root = i;
}
return Root;
}
void PrintLeaves(Tree R) //层序遍历
{
Queue Q;
Tree cur;
int count = 0;
if(R==Null) return;
Q=CreateQueue(QueueSize);
AddQ(Q, R);
while(!IsEmpty(Q)) {
cur = DeleteQ(Q);
if((T1[cur].Left == Null) && (T1[cur].Right == Null)) {
if(!count) {
printf("%d", cur);
count++;
}
else{
printf(" %d", cur);
continue;
}
}
if(T1[cur].Left!=Null) AddQ(Q, T1[cur].Left);
if(T1[cur].Right!=Null) AddQ(Q, T1[cur].Right);
}
}
int main()
{
Tree R;
R = BuildTree(T1);
PrintLeaves(R);
return 0;
}
3、03-树3 Tree Traversals Again (25分)
Sample Input:
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
Sample Output:
3 4 2 6 5 1
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
int count = 0;
typedef struct TreeNode *BinTree;
struct TreeNode {
int Data;
BinTree Left;
BinTree Right;
};
typedef struct SNode *Stack;
struct SNode {
BinTree Data;
Stack Next;
};
BinTree CreateBinTree(int data)
{
BinTree head;
head = (BinTree)malloc(sizeof(struct TreeNode));
head->Data = data;
head->Left = NULL;
head->Right = NULL;
return head;
}
int IsEmptyBinTree(BinTree BST)
{
return (BST == NULL);
}
void InsertLeftLeafe(BinTree BST, int leftData)
{
BinTree left;
left = (BinTree)malloc(sizeof(struct TreeNode));
left->Data = leftData;
left->Left = NULL;
left->Right = NULL;
BST->Left = left;
}
void InsertRightLeafe(BinTree BST, int rightData)
{
BinTree right;
right = (BinTree)malloc(sizeof(struct TreeNode));
right->Data = rightData;
right->Left = NULL;
right->Right = NULL;
BST->Right = right;
}
Stack CreateStack()
{
Stack S = (Stack)malloc(sizeof(struct SNode));
S->Data = NULL;
S->Next = NULL;
return S;
}
int IsEmptyStack(Stack S)
{
return (S->Next == NULL);
}
void StackPush(Stack S, BinTree pos)
{
Stack TmpCell = (Stack)malloc(sizeof(struct SNode));
TmpCell->Data = pos;
TmpCell->Next = S->Next;
S->Next = TmpCell;
}
BinTree StackPop(Stack S)
{
Stack FirstCell;
BinTree pos;
if(S->Next == NULL) {
printf("Stack Empty");
return NULL;
}
else {
FirstCell = S->Next;
S->Next = FirstCell->Next;
pos = FirstCell->Data;
free(FirstCell);
return pos;
}
}
void PrintStack(Stack S)
{
Stack head = S;
while(head) {
printf("%p ", head->Next);
head = head->Next;
}
printf("\n");
}
BinTree Read()
{
int N, data, count=0;
char str[10];
Stack S = CreateStack();
BinTree head;
BinTree BT = CreateBinTree(0);
head = BT;
scanf("%d\n", &N);
scanf("Push %d\n", &BT->Data);
StackPush(S, BT);
count++;
if(N) {
while(!IsEmptyStack(S)||count<N) { //是否堆栈为空，输入足够的结点
scanf("%s", str);
if(strcmp("Push", str) == 0) { //区分Push和Pop
scanf("%d", &data);
if(BT->Left == NULL) {
InsertLeftLeafe(BT, data);
BT = BT->Left;
StackPush(S, BT);
count++;
}
else if(BT->Right==NULL) {
InsertRightLeafe(BT, data);
BT = BT->Right;
StackPush(S, BT);
count++;
}
else {
printf("can't go here\n");
}
}
else {
BT = StackPop(S);
}
}
return head;
}
return NULL;
}
void PostOrderTraversal(BinTree BT){
if(BT){
PostOrderTraversal(BT->Left);
PostOrderTraversal(BT->Right);
if(!count) {
printf("%d", BT->Data);
count++;
}
else {
printf(" %d", BT->Data);
}
}
}
int main()
{
int N;
BinTree BT;
BT = Read();
PostOrderTraversal(BT);
return 0;
}
总结
简单总结下这周的学习内容，二叉树应该是数据结构的一大难点了，日常自闭，写了博客是为了能找着平板写一写代码，只能背一背了！！！
