03-Tree 3. Tree Traversals Again (25), 03-traversals

03-Tree 3. Tree Traversals Again (25), 03-traversals

Problem description:
The input displays non-recursive, stack operations in the middle order traversal, and post-order traversal in the output tree.
Key Issues
The push order is first traversed, And the pop order is central traversal.
The problem is resolved by first-order traversal and Middle-order traversal for subsequent traversal.

/*=============================================================================#       COPYRIGHT NOTICE#       Copyright (c) 2015##       @Author       :TonyChou#       @Email         :whzyb1991@163.com#       @Filename         :/home/zhoudaxia/program/PAT_MOOC_DS\03_3.cc#       @Last modified         :2015-06-30 21:22#       @Description    :=============================================================================*/#include<iostream>#include<stack>#include<vector>#include<string>#include<stdio.h>using namespace std;int find_head(vector<int> &v, int data){    for(int i = 0; i!= v.size(); ++i)        if(v[i]==data)            return i;}void find_post(vector<int> &v1, vector<int> &v2,int left2, int right2, int head1){    int head2 = find_head(v2, v1[head1]);    int left_size = head2 - left2;    if(head1!=v1.size()){    if(head2>left2)        find_post(v1,v2,left2, head2-1, head1+1);    if(head2<right2)        find_post(v1,v2,head2+1,right2,head1+left_size+1);    if(head1==0)        cout << v1[head1];    else        cout << v1[head1] <<" ";    }}int main(){    int N;    string s;    int node;    vector<int> pre_v;    vector<int> in_v;    stack<int> sta;    cin >> N;    for(int i=0;i != 2*N; ++i){        cin >> s;        if(s=="Push"){            cin>>node;            pre_v.push_back(node);            sta.push(node);                 }        else{            in_v.push_back(sta.top());            sta.pop();        }    }    find_post(pre_v,in_v,0,pre_v.size()-1,0);    return 0;}   

Analysis
First traverse in v1
1 2 3 4 5 6
Post-order traversal
3 2 4 1 6 5

The Recursive Parameter left2 right2 is the range of the Left subtree in v2, and head1 is the position of the header node in v1.
Head2 is the position of the header node in v2.
The exit of recursion is that head1 is traversed to the end of v1,
When head2! Recursion for left2 = left2, head2-1
Head2! = Right2 recursion [head2 + 1, right2]
Otherwise, v1 [head1] is output.

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