1086 Tree Traversals Again（中序 先序栈 构建 后序二叉树)

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:

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

 /**
本题题意：
    给定一个二叉树中序遍历栈， 现在给出所有结点入栈顺序， 和所有元素的出栈顺序，
本题思路：
     入栈的顺序是 先序遍历， 出栈的顺序是 中序遍历
     只不过题目没有说明 所有结点的值都是唯一的，因此
     设定 pre, in 数组 存储 下标索引, value数组存储对应的值
本题输入时 通过判断字符的长度 通过多组输入栈 s将输入的数字下标放入pre 与 in中
            因为多组输入， 由于命令窗没有接受到命令结束符， 因此命令窗口不能看到结果
      
**/

#include<iostream>
#include<stack>
#include<vector>
#include<cstring>
using namespace std;
vector<int> pre, post, in, value; //存放的下标索引位置 
stack<int> s;
void post_Order(int root, int start, int end){
	if(start > end)
		return;
	else{
		int i = start;
		while(i < end && in[i] != pre[root]) i++; //如果存在相同的数据 ， 此步就会出错(只会找第一个与 in[i]相等的值)， 因此这里存储的是下标 
												  //并不一定是根结点索引 
		post_Order(root + 1, start, i - 1);
		post_Order(root + i - start + 1, i + 1, end);
		post.push_back(pre[root]); 
	}
}
int main(){
	int n, num, index = 0;
	scanf("%d", &n);
	char c[6];
	while(~scanf("%s", c)){ //
		if(strlen(c) == 4){
			scanf("%d", &num);
			value.push_back(num);
			s.push(index);
			pre.push_back(index++);
		}else{
			in.push_back(s.top());
			s.pop();
		}
	}
	post_Order(0, 0, n - 1);
	for(int i = 0; i < n; i++){
		printf("%d", value[post[i]]);
		if(i != n - 1){
			printf(" ");
		}
	}
	return 0;
}

标签：1086,Again,int,Pop,pop,Push,二叉树,push,post
来源： https://blog.csdn.net/qq_41698081/article/details/99327455