1086 Tree Traversals Again

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：与上一题一样，也是先根据输入构造二叉树然后进行后序遍历进行输出。只是这道题没有明确的给出前序遍历和中序遍历，中序遍历根据题目的输出模仿栈的压入弹出就可以得到，前序遍历的话需要自己想一下，其实就是压栈的顺序就是前序遍历。

所以样例的前序遍历是：1 2 3 4 5 6，中序遍历是：3 2 4 1 6 5

#include<iostream>
#include<stdio.h>
#include<stack>
using namespace std;
struct node{
int left;
int right;
};
node tree[35];
int index = 0;
int pre[35];
int in[35];
void post(int *pre, int * in, int pBegin, int pEnd, int iBegin, int iEnd){
if (pBegin == pEnd || iBegin == iEnd){
return;
}
int iMid = 0;
for (int i = iBegin; i < iEnd; i++){
if (in[i] == pre[pBegin]){
iMid = i;
break;
}
}
int pMid = iMid - iBegin + pBegin + 1;
if (pMid > pBegin + 1){
tree[pBegin].left = pBegin + 1;
}
else{
tree[pBegin].left = -1;
}
if (pEnd > pMid){
tree[pBegin].right = pMid;
}
else{
tree[pBegin].right = -1;
}
post(pre, in, pBegin + 1, pMid, iBegin, iMid);
post(pre, in, pMid, pEnd, iMid + 1, iEnd);
}
void postShow(int root){
if (root == -1){
return;
}
postShow(tree[root].left);

postShow(tree[root].right);
if (index == 0){
printf("%d", pre[root]);
index++;
}
else{
printf(" %d", pre[root]);
}

}
int main(){
for (int n; scanf("%d", &n) != EOF;){
stack<int>temp;
int pIndex = 0;
int iIndex = 0;
for (int i = 0; i < n * 2; i++){
char op[6];
scanf("%s", op);
if (op[1] == 'u'){
scanf("%d", &pre[pIndex]);
temp.push(pre[pIndex]);
pIndex++;
}
else{
in[iIndex] = temp.top();
temp.pop();
iIndex++;
}
}
index = 0;
post(pre, in, 0, n, 0, n);
postShow(0);
printf("\n");
}
return 0;
}