PAT 1086 Tree Traversals Again

1086 Tree Traversals Again

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

 总结：经过前面几题树的遍历的问题，发现已知前序、中序转后续套路都是差不多的

（1）利用前序遍历或者后序遍历确定根节点

（2）利用中序遍历确定该根节点的左右子树

最后找到每一个点，从而可以获得一个序列

代码：

#include 
#include 
#include 
using namespace std;

vector pre,in,post;
void getpost(int root,int begin,int end){
    if(begin>end)   return;
    int i=begin;
    while(i<=end && in[i]!=pre[root]) i++;

    getpost(root+1,begin,i-1);
    getpost(root+(i-begin)+1,i+1,end);
    post.push_back(pre[root]);
}

int main(){
	stack s;
	int n;
	scanf("%d",&n);
	pre.resize(n),in.resize(n);
    
	int index1=0,index2=0;
	for(int i=0;i> t;
		
		if(t=="Push"){
			scanf("%d",&x);
			s.push(x);
			pre[index1++]=x;
		}
		else{
			in[index2++]=s.top();
			s.pop();			
		}
	}
    getpost(0,0,n-1);

    for(int i=0;i

好好学习，天天向上！

我要考研！