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

push为前序，pop为中序。

#include<stdio.h>#include<stdlib.h>#include<stack>using namespace std;int qx[35],zx[35];int cou1=0,cou2=0,flag=0;typedef struct Node *node;struct Node{int x;node left;node right;};node build(int s1,int e1,int s2,int e2){node T=(node)malloc(sizeof(struct Node));T->x=qx[s1];T->left=T->right=NULL;int root,i;for(i=s2;i<=e2;i++){if(zx[i]==qx[s1]){root=i;break;}}if(root!=s2){T->left=build(s1+1,root-s2+s1,s2,root-1);}if(root!=e2){T->right=build(root-s2+s1+1,e1,root+1,e2);}return T;}void post(node T){if(T){post(T->left);post(T->right);if(flag==0){printf("%d",T->x);flag=1;}else{printf(" %d",T->x);}}}int main(){int n,i,x;char str[5];scanf("%d",&n);stack<int>s;for(i=0;i<2*n;i++){scanf("%s",str);if(str[1]=='u'){scanf("%d",&x);qx[cou1++]=x;s.push(x);}else{zx[cou2++]=s.top();s.pop();}}node T=build(0,n-1,0,n-1); post(T);}