PAT 1086 Tree Traversals Again

1086Tree 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 integerN(≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 toN). Then2Nlines 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

注意点

• 这道题的本质是使用二叉树的前序序列和中序序列来构造出这棵二叉树，并对二叉树进行后序遍历。

• 对于输入的处理：

  • 字符串和数值分别处理
  • 使用栈

代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>

using namespace std;

const int maxn = 40;

int n;
int pre[maxn], in[maxn], post[maxn];
stack<int> st;
struct node
{
    int data;
    node* lchild, *rchild;
};

node* create(int preL, int preR, int inL, int inR)
{
    if (preL > preR)
        return 0;
    node* root = new node;
    root -> data = pre[preL];
    int k; 
    for (k = inL; k < inR; k ++)
    {
        if (in[k] == pre[preL])
            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()
{
//    freopen("test.txt", "r", stdin);
    scanf("%d", &n);
    char str[5];
    int x, preIndex = 0, inIndex = 0;
    for (int i = 0; i < 2 * n; i ++)
    {
        scanf("%s", str);
        if (strcmp(str, "Push") == 0)
        {
            scanf("%d", &x);
            pre[preIndex ++] = x;
            st.push(x);
        }
        if (strcmp(str, "Pop") == 0)
        {
            in[inIndex ++] = st.top();
            st.pop();
        }
    }
    node* root = create(0, n - 1, 0, n - 1);
    postOrder(root);
    
    return 0;
}