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.
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
題目大意
樹的遞歸遍歷可以借助堆棧變成非遞歸遍歷，本題提供非遞歸過程中堆棧操作順序堡掏，以此構(gòu)建二叉樹报亩，并求出這個(gè)二叉樹的后序。
分析
本題「新瓶裝舊酒」盈匾，本質(zhì)依然是考察已知樹的兩種遍歷序列求其他遍歷序列，這個(gè)問題我已經(jīng)在一篇博客專門詳盡闡述了「知兩序求兩序」這類問題的解法[1]毕骡，此題只需將問題轉(zhuǎn)化為「知兩序求兩序」即可解出削饵，觀察可知，push的數(shù)為前序序列未巫，pop的數(shù)為中序序列窿撬。
#include <iostream>
#include <vector>
#include <stack>
using namespace std;
vector<int> pre,in,post;
stack<int> st;
void post_travel(int root,int start,int end){
if(start>end) return;
int i=start;
while(i<=end && pre[root]!=in[i]) i++;
post_travel(root+1,start,i-1);
post_travel(root+1+i-start,i+1,end);
post.push_back(pre[root]);
}
int main(){
int n;
cin>>n;
for(int i=0;i<n*2;i++){
string s;
int val;
cin>>s;
if(s=="Push"){
cin>>val;
pre.push_back(val);
st.push(val);
}else if(!st.empty()){
in.push_back(st.top());
st.pop();
}
}
post_travel(0,0,n-1);
for(int i=0;i<n;i++){
if(i==0) cout<<post[i];
else cout<<" "<<post[i];
}
return 0;
}
