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
解題思路
通過stack模擬， push表示二叉樹的前序值，pop表示二叉樹的中序的值。 完成輸入後則得到中序和前序遍歷結果。通過中序和前序構建二叉樹（建立的過程類似於1020）之後，後序遍歷輸出即可。
解題代碼
#include <iostream>
#include <stack>
#include <cstdio>
using namespace std;
int n, lenp, leni, cnt, pre[30], in[30];
string op;
stack<int> st;
struct node{
int data;
node* lchild;
node* rchild;
};
void post(node* root){
if (root == NULL) return;
post(root->lchild);
post(root->rchild);
printf("%d", root->data);
cnt ++;
if (cnt < n) printf(" ");
}
node* create(int preL, int preR, int iL, int iR){
if (preL > preR) return NULL;
node* root = new node;
root->data = pre[preL];
int k;
for (k = iL ; k < iR; k++)
if (in[k] == pre[preL]) break;
int numberL = k - iL;
root->lchild = create(preL + 1, preL + numberL, iL, k - 1);
root->rchild = create(preL + numberL + 1, preR, k + 1, iR);
return root;
}
int main(){
scanf("%d", &n);
for (int i = 0; i < 2 * n; i++){
cin >> op;
if (op == "Push"){
int t;
scanf("%d",&t);
pre[lenp++] = t;
st.push(t);
} else{
in[leni++] = st.top();
st.pop();
}
}
node* root = create(0, n - 1, 0, n - 1);
post(root);
return 0;
}