1086 Tree Traversals Again-PAT甲级

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本文介绍如何使用栈的push和pop操作序列生成并遍历一棵二叉树，通过给定的节点push顺序（先序遍历）和pop顺序（中序遍历），可以唯一确定一棵二叉树，并实现其后序遍历。

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

stack中的push顺序为二叉树先序遍历的结果，stack中的pop的顺序为二叉树中序遍历的结果，可以根据二叉树先序、中序遍历的结果来重建二叉树，进而进行后序遍历

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=35;
struct node{
	int v;//数据域
	node *left=NULL;
	node *right=NULL; 
};
int pre[N],in[N];//存储中序遍历的结果
int n;//结点的个数
//先序+中序建树的过程 
node* create(int prel,int prer,int inl,int inr){
	if(prel>prer){
		return NULL;
	}
	node *root=new node;
	root->v=pre[prel];
	int k;
	for(k=inl;k<=inr;k++){
		if(in[k]==pre[prel])
			break;
	}
	int numleft=k-inl;
	root->left=create(prel+1,prel+numleft,inl,k-1);
	root->right=create(prel+numleft+1,prer,k+1,inr);
	return root;	
} 
int flag=0;
//后序遍历的过程
void postorder(node *root){
	if(root==NULL) return;
	postorder(root->left);
	postorder(root->right);
	if(flag) printf(" ");
	printf("%d",root->v);
	flag=1;
} 
int main(){
	scanf("%d",&n);
	int pre_num=-1,in_num=-1;
	string s;
	int x;
	stack<int>st;
	for(int i=1;i<=2*n;i++){
		cin>>s;
		if(s=="Push"){//push的输入为树的先序，pop的输出为树的中序 
			cin>>x;
			st.push(x);
			pre_num++;
			pre[pre_num]=x;
		}else{
			in_num++;
			in[in_num]=st.top();
			st.pop();
		}	
	}
	node *root=create(0,pre_num,0,in_num);
	postorder(root);
	return 0;
}