1086. Tree Traversals Again (25)

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

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

思路：

由题意可得出，入栈是用的先序遍历，出栈用的是中序遍历。然后利用先序序列和中序序列，重构二叉树，并用后序遍历输出二叉树。

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<stack> 
using namespace std;


struct Node{
	int num;
	struct Node *left;
	struct Node *right;
};
typedef struct Node *TNode;

int pre[31],in[31],post[31];
int n;

TNode create(int preL,int preR,int inL,int inR)/*当前的二叉树，[preL,preR]为先序序列区间，[inL,inR]为中序序列*/
{
	if(preL > preR)/*先序序列长度<0,则返回*/
	{
		return NULL;
	}
	
	TNode r = (TNode)malloc(sizeof(struct Node));
	r->num = pre[preL];
	r->left = NULL;
	r->right = NULL;
	
	int k;
	for(k = inL; k <= inR; k++)
	{
		if(in[k] == pre[preL])break;
	}
	
	int numleft = k - inL;/*记录当前根结点左子树的个数*/
	
	r->left = create(preL+1,preL+numleft,inL,k-1);/*返回左子树的根结点地址，并赋值给r的左指针*/
	r->right = create(preL+numleft+1,preR,k+1,inR);/*返回右子树的根结点地址，并赋值给r的右指针*/

	return r;
	
}
int num = 0;
void postorder(TNode r)/*后序遍历输出*/
{
	if(r == NULL)
	{
		return;
	}
	postorder(r->left);
	postorder(r->right);
	printf("%d",r->num);
	num++;
	if(num < n)
	{
		printf(" ");
	}
}

int main(void)
{
	scanf("%d",&n);
	int j = 0,k = 0;
	int number;
	char str[5];
	stack<int> st;
	for(int i = 0; i < 2*n; i++)
	{
		scanf("%s",str);
		getchar();
		if(strcmp(str,"Push") == 0)
		{
			scanf("%d",&number);
			pre[j++] = number;
			st.push(number);
		}
		else
		{
			in[k++] = st.top();
			st.pop();
		}
	}
	TNode r = create(0,n - 1,0,n - 1);
	postorder(r);
	
	return 0;
}