【MOOC】03-树3 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

通过观察可以发现，该题实际上是给出二叉树的前序和中序遍历，然后输出该二叉树的后序遍历。测试样例一直过不去，后来发现是n-i-1，不是n-i+1，注意每次递归时下标的变化。

说明：bef代表before（前序起始），mid代表in（中序起始），aft代表after（后序起始）。pre为前序遍历，in为中序遍历，post为后序遍历

#include <iostream> 
#include <stack> 
using namespace std;

int pre[35],in[35],post[35];

void order(int bef,int mid,int aft,int n){
	if(n<1)
		return;
	if(n==1)//递归终止条件
		post[aft] = pre[bef];
	else{
		int tmp = pre[bef];
		post[aft+n-1] = tmp;
		int i;
		for(i = 0;i<n;i++){
			if(in[mid+i]==tmp)
				break;
		}
		order(bef+1,mid,aft,i);
		order(bef+i+1,mid+i+1,aft+i,n-i-1);
	}
}

void print(int n){
	if(n<1)
		return;
	cout << post[0];
	for(int i = 1;i<n;i++)
		cout << " " << post[i];
}

int main(){
	int n,x,m;
	cin >> n;
	stack<int> s;
	int k = 0,p = 0;
	string op;
	//注意是2*n，因为n个结点的中序遍历对应2*n次出栈和入栈
	for(int i = 0;i<2*n;i++){
		cin >> op;
		if(op=="Push"){
			cin >> x;
			pre[k++] = x;
			s.push(x);
		}
		else if(op=="Pop"&&!s.empty()){
			m = s.top();
			s.pop();
			in[p++] = m;
		}
	}
	order(0,0,0,n);
	print(n);
	return 0;
}