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:

6Push 1Push 2Push 3PopPopPush 4PopPopPush 5Push 6PopPop

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;}