PAT甲级1086 Tree Traversals Again (25 分)

最新推荐文章于 2024-09-21 22:25:33 发布

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最新推荐文章于 2024-09-21 22:25:33 发布

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本文探讨了使用栈进行二叉树中序遍历的方法，分析了如何从栈的推入和弹出操作序列生成唯一二叉树，并给出将前序和中序遍历转换为后序遍历的解决方案。

1086 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.

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:

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

题目大意：

用栈来代替递归中序遍历树。输出后序遍历序列。

思路：

有个很重要的性质，入栈顺序为前序序列，出栈顺序为中序序列。懂得这个性质则这道题就是paper tiger。

实质上就是在问如何前序 + 中序转后序。

前序 + 中序转后序的详细分析：https://blog.youkuaiyun.com/qq_42306885/article/details/99021038

参考代码：

#include<stack>
#include<string>
#include<vector>
#include<iostream>
using namespace std;
vector<int> in, pre;
stack<int> st;
string s;
int n, temp, cnt;
void post(int root, int start, int end){
	if(start > end)	return;
	int i = start;
	while(i < end && in[i] != pre[root]) i++;
	post(root + 1, start, i - 1);
	post(root + i - start + 1, i + 1, end);
	if(cnt == 0){
		cnt++;
		printf("%d", pre[root]);
	}else	printf(" %d", pre[root]);
}
int main(){
	scanf("%d", &n);
	for(int i = 0; i < 2 * n; ++i){
		cin >> s;
		if(s.size() == 4){
			cin >> temp;
			st.push(temp);
			pre.push_back(temp);
		}else{
			in.push_back(st.top());
			st.pop();
		}
	}
	post(0, 0, n - 1);
	return 0;
}