03-树3 Tree Traversals Again (25分)

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integerNN ($\le 30$) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 toNN). Then $2N$ lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

Sample Output:

3 4 2 6 5 1

思路：

1、入栈过程是树的先序遍历，出栈过程是树的中序遍历

2、已知先序和中序，求后序；

参考：已知前序（先序）与中序输出后序

#include <stdio.h>#include <stdlib.h>int *pre, *in;int flag = 0;//栈的结构数组表示#define Error -1;typedef int Position;typedef int ElementType;typedef struct SNode *Stack;struct SNode {ElementType * Data;Position Top;int MaxSize;};Stack CreateStack( int MaxSize ){Stack S = (Stack)malloc(sizeof(struct SNode));S->Data = (ElementType *)malloc(MaxSize * sizeof(ElementType));S->Top = -1;S->MaxSize = MaxSize;return S;}bool IsFull( Stack S ){return S->Top == S->MaxSize - 1; }bool Push( Stack S, ElementType X){if( IsFull(S) ) {printf("stack is full");return false;}else {S->Data[++(S->Top)] = X;return true;}}bool IsEmpty( Stack S ){return S->Top == -1;}ElementType Pop( Stack S ) {if( IsEmpty(S) ){printf("stack is empty");return Error;}else {return S->Data[S->Top--];}}void StackClear( Stack S ){S->Top = -1;}void Post(int root, int start, int end){//当start = end时，还可以继续做下去。（具体细节未细究）if(start > end) return ;//i - start 是左子树的结点个数，当start = 0， i是左子树结点个数int i = start;while(i < end && in[i] != pre[root]) i++; //试下i <= end; 结果：可以。//原因：因为一定会有根，当i = end时，后面的条件仍会使循环结束//（所以也就是前面的条件使循环结束，还是后面的条件使循环结束的问题了）//start和end 是按 in 中的下标确定的， root是由 pre 中的下标确定(即 pre 的作用是 “寻根”)//左子树递归；//root + 1 是左子树的根（根左右），//i是根， i-1就是左子树的最后一个元素(左根右）Post( root + 1, start, i - 1 );//右子树递归//root + 1 + i - start 是右子树的根（根左右），因此要加上 i - start（左子树的结点个数）//i是根， i+1就是右子树的第一个元素(左根右）Post( root + 1 + i - start, i + 1, end); if(!flag){printf("%d", pre[root]);flag = 1;}else printf(" %d", pre[root]);}int main(){int N, i = 0, j = 0, val;char s[10];Stack S;scanf("%d", &N);S = CreateStack(N);pre = (int *)malloc(N * sizeof(int));in = (int *)malloc(N * sizeof(int));for(int k = 0; k < 2 * N; k++){    //不能用i，会和if语句中的i混淆，错误原因scanf("%s", s);//printf("%s\n", s);if( s[1] == 'u' ){scanf("%d", &val);pre[i++] = val;Push(S, val);}else{in[j++] = Pop(S);}}/*for(int i = 0; i < N; i++)printf("%d ", pre[i]);printf("\n");for(int i = 0; i < N; i++)printf("%d ", in[i]);*/Post(0, 0, N - 1);system("pause");return 0;}