1086 Tree Traversals Again (25分)

1 题目

1086 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 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

Sample Output:
3 4 2 6 5 1

2 解析

• 题意：根据栈的出栈、入栈序列，输出二叉树的后序序列
• 思路：
  • 入栈序列为先序序列，出栈序列为中序序列
  • 1 根据先序和中序序列建立二叉树，然后后序遍历输出二叉树

3 参考代码

#include <cstdio>
#include <iostream>
#include <stack>
#include <string>

using std::cin;
using std::stack;
using std::string;

const int MAXN = 35;
int in[MAXN];
int pre[MAXN];
int N;

struct node
{
    int data;
    node* lchild;
    node* rchild;
};

node* Create(int preL, int preR, int inL, int inR){
    if(preL > preR){
        return NULL;
    }

    node* root = new node;
    root->data = pre[preL];

    int k;
    for (k = inL; k <= inR; ++k)
    {
        if(in[k] == root->data){
            break;
        }
    }

    int numLeft = k - inL;

    root->lchild = Create(preL + 1, preL + numLeft, inL, k - 1);
    root->rchild = Create(preL + numLeft + 1, preR, k + 1, inR);

    return root;
}

int num = 0;
void postorder(node* root){
    if(root == NULL){
        return;
    }

    postorder(root->lchild);
    postorder(root->rchild);

    printf("%d", root->data);
    num++;
    if(num < N) printf(" ");
}

int main(int argc, char const *argv[])
{

    scanf("%d", &N);

    string str;
    int temp, num,inCount = 0,preCount = 0;
    stack<int> s;
    for (int i = 0; i < 2 * N; ++i)
    {
        cin >> str;
        if(str == "Push"){
            scanf("%d", &num);
            s.push(num);
            pre[preCount++] = num;
        }else{
            temp = s.top();
            s.pop();
            in[inCount++] = temp;
        }
    }

    node* root = Create(0, N - 1, 0, N - 1);

    postorder(root);

    return 0;
}