7-92 Tree Traversals Again

题目如下：

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

用父结点来回归。直到父结点为左子树时，才赋值（可联想中序遍历访问的前提）。代码1如下：

#if 1
#include<cstdio>
#include<cstring>
using namespace std; 
int const int maxn=3e+5;
int pa[maxn],ld[maxn],rd[maxn];
int flag=0;
void dfs(int u)
{
	
	if(ld[u]!=-1)
	dfs(ld[u]);
	if(rd[u]!=-1)
	dfs(rd[u]);
	if(flag==0)
	{
		printf("%d",u);
		flag=1;
	}
	else
	  printf(" %d",u);
 } 
int main()
{
	int n,u;
	char ope[5];
	scanf("%d%s%d",&n,ope,&u);
	for(int i=1;i<=n;i++)
		pa[i]=ld[i]=rd[i]=-1;
	int t=u,v=u,flag=1; 
	for(int i=0;i<n*2-1;i++)
	{
		scanf("%s",ope);
		if(strcmp(ope,"Push")==0)
		{
			scanf("%d",&u);
			if(flag)
				ld[v]=u;
			else
				rd[v]=u;
			pa[u]=v;
  			v=u; 
  			flag=1;
		}
		else
		{
		  	if(!flag) //若已经出过一次栈了
		  	{
		    	    while(ld[pa[v]]!=v)//直到为左子树
                                v=pa[v];
		    	    v=pa[v];//才算回归
		  	}
			flag=0;
		}
	}
	dfs(t);
	printf("\n");
	return 0;
}
#endif

用栈来回归。由于只入一次栈只出一次栈，所以效果和中序递归是一样的。代码2如下：

#if 1
#include<cstdio>
#include<cstring>
#include<stack> 
using namespace std; 
int const int maxn=3e+5;
int ld[maxn],rd[maxn];
int flag=0;
void dfs(int u)
{
	
	if(ld[u]!=-1)
	dfs(ld[u]);
	if(rd[u]!=-1)
	dfs(rd[u]);
	if(flag==0)
	{
		printf("%d",u);
		flag=1;
	}
	else
	  printf(" %d",u);
 } 
int main()
{
	int n,u;
	char ope[5];
	stack<int>S;
	scanf("%d%s%d",&n,ope,&u);
	for(int i=1;i<=n;i++)
		ld[i]=rd[i]=-1;
        S.push(u);
	int t=u,v=u,flag=1; 
	for(int i=0;i<n*2-1;i++)
	{
		scanf("%s",ope);
		if(strcmp(ope,"Push")==0)
		{
			scanf("%d",&u);
			if(flag)
				ld[v]=u;
			else
				rd[v]=u;
  		        v=u; 
  		        S.push(v);
  		        flag=1;
		}
		else
		{
			v=S.top();
			S.pop();
			flag=0;
		}
	}
	dfs(t);
	printf("\n");
	return 0;
}
#endif

代码1，2均为前序、中序的混合模拟，得出树的结构之后，进行深度搜索即可。以上。