1086. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

push 的顺序就是二叉树的前序

pop的顺序就是二叉树的中序遍历

本质上还是考根据这两个顺序建立二叉树，并且进行后序遍历

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

来源： <http://www.patest.cn/contests/pat-a-practise/1086>

#include<iostream>
#include<vector>
#include<string>
#pragma warning(disable:4996)

using namespace std;
vector<int> s;
int inOrder[32] = { 0 },preOrder[32] = { 0 };
int n;

struct Node {
	int val;
	Node* left = NULL;
	Node* right = NULL;
};

Node* BuildTree(int* pre, int* in, int n) {
	if (n <= 0)
		return NULL;
	int i;
	for (i = 0; i < n; i++) {
		if (*(in + i) == *pre) {
			break;
		}
	}
	Node* p = (Node*)malloc(sizeof(Node));
	p->val = *pre;
	p->left = BuildTree(pre + 1, in, i);
	p->right = BuildTree(pre + i + 1, in + i + 1, n - i - 1);

	return p;
}

void PostOrder(Node *p) {
	if (p->left != NULL)
		PostOrder(p->left);
	if (p->right != NULL)
		PostOrder(p->right);
	if (p->val != preOrder[0])
		cout << p->val << " ";
	else
		cout << p->val;
}

int main(void) {
	freopen("Text.txt", "r", stdin);

	cin >> n;
	string str;
	for (int i = 0; i < 2 * n; i++) {
		cin >> str;
		if (str == "Push") {
			int temp;
			cin >> temp;
			for (int j = 0; j < 31; j++) {
				if (preOrder[j] == 0) {
					preOrder[j] = temp;
					break;
				}
			}
			s.push_back(temp);
		}
		else {
			for (int j = 0; j < 31; j++) {
				if (inOrder[j] == 0) {
					inOrder[j] = s[s.size() - 1];
					break;
				}
			}
			s.pop_back();
		}
	}

	Node* root = (Node*)malloc(sizeof(Node));

	root = BuildTree(preOrder, inOrder, n);

	PostOrder(root);

	return 0;
}

来自为知笔记(Wiz)