1086. Tree Traversals Again (25)【二叉树】——PAT (Advanced Level) Practise

题目信息

1086. Tree Traversals Again (25)

时间限制200 ms

内存限制65536 kB

代码长度限制16000 B

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

解题思路

二叉树遍历转换

AC代码

#include <cstdio>
#include <stack>
using namespace std;

int preorder[35], inorder[35];
int n, preid = 0, inid = 0, cnt = 0;
int get(){
    char s[10];
    scanf("%s", s);
    if (s[1] == ‘o‘) return -1;
    int a;
    scanf("%d", &a);
    return a;
}
void build(int preb, int pree, int inb, int ine){
    if (preb > pree) return;
    int root = preorder[preb];
    int inroot = inb;
    while (inorder[inroot] != root) ++inroot;
    build(preb+1, preb+inroot-inb, inb, inroot-1);
    build(preb+inroot-inb+1, pree, inroot+1, ine);
    if (cnt++ != 0) putchar(‘ ‘);
    printf("%d", root);
}
int main(){
    scanf("%d", &n);
    stack<int> st;
    for (int i = 0; i < n*2; ++i){
        int a = get();
        if (a != -1){
            st.push(a);
            preorder[preid++] = a;
        }else{
            inorder[inid++] = st.top();
            st.pop();
        }
    }
    build(0, n-1, 0, n-1);
    return 0;
}