PAT A1086 Tree Traversals Again (25分)

原题

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

大体题意

中序遍历可以用栈进行非递归遍历，因此要求通过栈的活动得到树，并输出后序序列

思路

前序序列可以看成入栈
中序序列可以看成出栈
因此通过所给的输入，我们可以得到前序序列和中序序列
因此该题的本质就是通过前序和中序得到后序

代码

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

vector<int> preorder, inorder;
stack<int>node;

struct Tree
{
	int data;
	Tree* lchild, * rchild;
};

Tree* createTree(int preL, int preR, int inL, int inR)
{
	if (inL > inR)
		return NULL;
	Tree* t = new Tree;
	t->data = preorder[preL];

	int i;
	for (i = inL; i < inR; ++i)
		if (inorder[i] == preorder[preL])
			break;

	int numleft = i - inL;
	t->lchild = createTree(preL + 1, preL + numleft, inL, i - 1);
	t->rchild = createTree(preL + numleft + 1, preR, i + 1, inR);
	return t;
}

bool first = false;
void postTravel(Tree* root)
{
	if (root == NULL)
		return;
	postTravel(root->lchild);
	postTravel(root->rchild);
	if (first)
		cout << " ";
	cout << root->data;
	first = true;
}

int main()
{
	int n, a;
	cin >> n;
	string s;
	for (int i = 0; i < 2 * n; i++)
	{
		cin >> s;
		if (s == "Push")
		{
			cin >> a;
			preorder.push_back(a);
			node.push(a);
		}
		else
		{
			a = node.top();
			node.pop();
			inorder.push_back(a);
		}
	}

	Tree* root = createTree(0, n - 1, 0, n - 1);
	postTravel(root);
	return 0;
}

运行结果