PAT 1086. Tree Traversals Again (25)

1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

思路：根据根左右和左根右的顺序，构造树，再用左右根的顺序遍历

#include <stdio.h>
#include <string.h>
#include <vector>
using namespace std;
int N, X, pre[50], in[50];
char op[10];
vector<int> list, post_order;
struct Node
{
	int val;
	Node *left, *right, *root;
};

Node * fun(int *pre_order, int *in_order, int length)
{
	if(length == 0)
	{
		return NULL;
	}
	int val = pre_order[0], index = 0;
	for(int i = 0; i < length; i++)
	{
		if(pre_order[0] == in_order[i])
		{
			index = i;
			break;
		}
	}
	Node *newNode = new Node();
	newNode->val = pre_order[0];
	newNode->left = fun(pre_order + 1, in_order, index);
	newNode->right = fun(pre_order + index + 1, in_order + index + 1, length - 1 - index);
	return newNode;
}

void post(Node *root)
{
	if(root == NULL)
	{
		return;
	}
	post(root->left);
	post(root->right);
	post_order.push_back(root->val);
}

int main()
{
	while(scanf("%d", &N) != EOF)
	{
		Node *root;
		int index_pre = 0, index_in = 0;
		for(int i = 0; i < N * 2; i++)
		{
			scanf("%s", &op);
			if(strcmp(op, "Push") == 0)
			{
				scanf("%d", &X);
				pre[index_pre++] = X;
				list.push_back(X);
			}
			else
			{
				in[index_in++] = list.back();
				list.pop_back();
			}
		}
		root = fun(pre, in, index_pre);
		post(root);
		printf("%d", post_order[0]);
		for(int i = 1; i < index_pre; i++)
		{
			printf(" %d", post_order[i]);
		}
		printf("\n");
	}
	return 0;
}