PAT 甲级 1086 Tree Traversals Again (25分)（先序中序链表建树，求后序）***重点复习

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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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：

用栈的形式给出一棵二叉树的建立的顺序，求这棵二叉树的后序遍历

题解：

栈实现的是二叉树的中序遍历（左根右），而每次push入值的顺序是二叉树的前序遍历（根左右），所以该题可以用二叉树前序和中序转后序的方法做～

AC代码：

#include<bits/stdc++.h>
using namespace std;
int n;
struct node{
    int data;
    node *left,*right;
};
vector<int>pre,in,post;
stack<int>s;
node *buildTree(vector<int>pre,vector<int>in,int pl,int pr,int il,int ir){
    if(pl>pr || il>ir) return NULL;
    int pos=-1;
    for(int i=il;i<=ir;i++){
        if(in.at(i)==pre.at(pl)){
            pos=i;
            break;
        }
    }
    node *root=new node();
    //root->left=root->right=NULL;
    root->data=pre.at(pl);
    root->left=buildTree(pre,in,pl+1,pl+pos-il,il,pos-1);
    root->right=buildTree(pre,in,pl+pos-il+1,pr,pos+1,ir);
    return root;    
}
void postorder(node *root){
    if(root){
        postorder(root->left);
        postorder(root->right);
        post.push_back(root->data);    
    }
}
int main(){
    cin>>n;
    pre.push_back(-1);
    in.push_back(-1);
    char c[10];
    int x;
    for(int i=1;i<=2*n;i++){
        cin>>c;
        if(strcmp(c,"Push")==0){
            cin>>x;
            s.push(x);
            pre.push_back(x);
        }else{
            in.push_back(s.top());
            s.pop();
        }
    }
    node *root = buildTree(pre,in,1,n,1,n);
    postorder(root);
    for(int i=0;i<post.size();i++){
        cout<<post.at(i);
        if(i!=post.size()-1) cout<<" ";
    }
    return 0;
}