1086. Tree Traversals Again （树的遍历，前序中序转后序）

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

题目大意：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

需要一个堆结构s，一个child变量（表示该节点是其父亲节点的左孩子还是右孩子），父亲节点fa
对于push v操作：
1).第一个push肯定是根节点root。
2).根据child变量，建立fa与v的父子关系。
3).由于是中序遍历，所以接下来的节点必定是v的left（如果有的话），child=left,fa=v;
4).然后进行push操作

对于pop操作：
1).根据中序遍历性质，可知接下来的节点必定是pop节点的右孩子（如果有的话）,child=right,fa=s.top()
2).进行pop操作。

#include<iostream>
#include<algorithm>
#include<string.h>
#include<stack>
#define LEFT 0
#define RIGHT 1
#define maxn 100
using namespace std;
stack<int> s;
//着重注意这题考查的是"建树"的技巧 
struct  NODE
{
   int left=-1;	
   int right=-1;	 
}node[maxn];

bool first=true;
void postOrder(int u){//虽然建树时是按中序建的树
    if(u==-1)         //但是按后序扫的
        return;
    postOrder(node[u].left);
    postOrder(node[u].right);
    if(first){
        first=false;
        printf("%d",u);
    }
    else{
        printf(" %d",u);
    }
}

int main()
{
	int n,v;
	int root=-1,fa;
	//fa是为了记录当前结点
	//以便把各个node结点相连 
	int child=LEFT;
	//child变量记录的是,
	//当前父节点往左走，还是往右走 
	char str[10];
	scanf("%d",&n);
	while(scanf("%s",str)!=EOF)
	{
		if(strcmp(str,"Push")==0)
		{
			scanf("%d",&v);
			if(root==-1)//为这棵树找根节点，入口 
			root=v;
			else 
			{
				if(child==LEFT)
                node[fa].left=v;
				else node[fa].right=v;				
			}
			fa=v;//更新父节点 
			child=LEFT;
			s.push(v); 
		}//对于入栈的操作,
		 //由于是中序，新节点优先进左子树 
		else//出栈后在改变方向,
		    //再次push时向右子树push 
		{
			child=RIGHT;
			fa=s.top(); 
			s.pop();
		}
	}
	postOrder(root);
    return 0;	
}

1086. Tree Traversals Again （树的遍历，前序中序转后序）

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