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
题意:根据中序前序,求出后序。
这道题能看懂样例,就很容易写出来,关键点是所有的入栈Push元素顺序构成的序列为树的前序遍历序列;所有的出栈Pop元素顺序构成的序列为中序遍历序列。
//根据 中序 前序 找 后序
#include <bits/stdc++.h>
using namespace std;
vector<int> pre, in, post;
int n, num;
string opt;
//rt是前序当前根节点的下标
//st,en 分别是当前子树在中序序列最左端和最右端的下标
void postOrder(int rt, int st, int en){
if(st > en) return ;
int i = st;
while(i < en && in[i] != pre[rt]) i++;
postOrder(rt + 1, st, i - 1);
postOrder(rt + 1 + i - st, i + 1, en);
post.push_back(pre[rt]);
}
int main(){
stack<int> s;
while(!s.empty()) s.pop();
cin >> n;
for(int i = 0; i < 2*n; i++){
cin >> opt;
if(opt == "Push"){ //前序
cin >> num;
s.push(num);
pre.push_back(num);
}else{ //中序
in.push_back(s.top());
s.pop();
}
}
postOrder(0, 0, n-1);
for(int i = 0; i < n; i++){
cout << post[i] ;
if( i != n-1) cout << " ";
}
cout << endl;
return 0;
}