題目表述

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.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). 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:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
結尾無空行
 

Sample Output:

3 4 2 6 5 1
結尾無空行

題目解析

本題借鑑了一些演算法：通過輸入可以構建前序遍歷陣列，和中序遍歷陣列，然後通過遞迴，根據前項遍歷陣列和中序遍歷陣列得出後續遍歷陣列。
1.首先棧push的過程即是前序遍歷的陣列；
pop的過程即是中序遍歷的陣列。
2.後序遍歷：將前序的第一個放到後序的最後一個，然後通過中序得到左節點數L和右節點數R，將前序的下一個放到後序陣列的左邊的最後一個，右邊同，然後遞迴實現。（每次遞迴前序陣列後移一個，中序用來找左數L和右數R，將前序的本節點賦值給後序陣列的L或R個元素的最後一個）

程式碼

// Tree traversal again
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define Max 30
int pre[Max], in[Max], post[Max];

typedef struct node {
    int data;
    struct node* next;
} Node, *Stack;
typedef struct tnode {
    int data;
    struct tnode *lchild, *rchild;
} TNode, *Tree;

//函式宣告
Stack Init_stack();
void push(Stack S, int x);
bool pop(Stack S, int* x);
bool isempty(Stack S);
void buildTree(int N);
void PostTree(int pre_r, int in_r, int post_r, int N);

//主函式
int main() {
    int N;
    scanf("%d", &N);
    buildTree(N);
    PostTree(0, 0, 0, N);
    for (int i = 0; i < N; i++) {
        printf("%d", post[i]);
        if (i < N - 1)
            putchar(' ');
    }
    return 0;
}

//函式定義
Stack Init_stack() {
    Stack p = (Stack)malloc(sizeof(Node));
    p->next = NULL;
    return p;
}
void push(Stack S, int x) {
    Node* p = (Node*)malloc(sizeof(Node));
    p->data = x;
    p->next = S->next;
    S->next = p;
}
bool pop(Stack S, int* x) {
    if (isempty(S))
        return false;
    Node* p = S->next;
    *x = p->data;
    S->next = p->next;
    free(p);
    return true;
}
bool isempty(Stack S) {
    if (S->next == NULL)
        return true;
    else
        return false;
}

void buildTree(int N) {
    Stack S = Init_stack();

    int index_in = 0;   //中序陣列角標
    int index_pre = 0;  //前序陣列角標
    for (int i = 0; i < 2 * N; i++) {
        char order[5];
        int data;
        scanf("%s", order);
        if (strcmp(order, "Push") == 0) {
            scanf("%d", &data);
            pre[index_pre++] = data;
            push(S, data);
        } else {
            pop(S, in + index_in);
            index_in++;
        }
    }
    return;
}
void PostTree(int pre_r, int in_r, int post_r, int N) {
    if (N == 0)
        return;
    if (N == 1)
        post[post_r] = pre[pre_r];
    int i;
    int root = pre[pre_r];
    post[post_r + N - 1] = root;
    for (i = 0; i < N; i++) {
        if (in[in_r + i] == root)
            break;
    }
    int L = i, R = N - i - 1;
    PostTree(pre_r + 1, in_r, post_r, L);
    PostTree(pre_r + 1 + L, in_r + L + 1, post_r + L, R);
}