03-树3 Tree Traversals Again (25分)

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 $N$ ($\le 30$) 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

这题参照王道考研机试指南P53。

push为前序,pop为中序,然后建树.后序输出

#include <stack>//模拟 王道考研机试指南 #include <string.h>#include<stdio.h>using namespace std;typedef struct Node *node;struct Node{node left;node right;int num;}tree[30];int count=0,cou3=0;int qx[30],zx[30],hx[30];node chuli(int s1,int e1,int s2,int e2){tree[count].left=tree[count].right=NULL;tree[count].num=qx[s1];node head=&tree[count++];int root,i;for(i=s2;i<=e2;i++){if(zx[i]==qx[s1]){root=i;break;}}if(root!=s2){head->left=chuli(s1+1,root-s2+s1,s2,root-1);}if(e2!=root){head->right=chuli(root-s2+s1+1,e1,root+1,e2);}return head;}void post(node T){if(T){post(T->left);post(T->right);hx[cou3++]=T->num;}}int main(){int n,i;scanf("%d",&n);int cou1=0,cou2=0;char str[10];int temp;stack<int>s;for(i=0;i<2*n;i++){scanf("%s",str);if(strlen(str)==4){scanf("%d",&temp);qx[cou1++]=temp;s.push(temp);}else{zx[cou2++]=s.top();s.pop();}}node head=chuli(0,n-1,0,n-1);post(head);printf("%d",hx[0]);for(i=1;i<n;i++){printf(" %d",hx[i]);}}