Tree Traversals Again (25)

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

分析：

題目通過棧的方式給出二叉樹的周遊過程，Push 就是先序周遊，Pop就是中序周遊。

可以開二個數組

作為儲存兩個序列 pre[maxn] ino[maxn]

然後根據兩個序列來構造二叉樹（遞歸的方式）

注意終止條件： if(preL>preR) return NULL;

以及各個邊界

#include<bits/stdc++.h>
using namespace std;
const int maxn=;
int pre[maxn],ino[maxn],post[maxn];
struct node
{
  int data;
  node *lchild;
  node *rchild;
};
node* create(int preL,int preR,int inoL,int inoR)
{
    if(preL>preR) return NULL;
    node *root= new node;
    root->data=pre[preL];
    int k;
    for(k=inoL;k<=inoR;k++)
    {
        if(ino[k]==pre[preL])
            break;
    }
    int numleft=k-inoL;
    root->lchild=create(preL+,preL+numleft,inoL,k-);
    root->rchild=create(preL+numleft+,preR,k+,inoR);
    return root;
}
int num=,n;
void postorder(node *root)
{
     if(root==NULL) return;
     postorder(root->lchild);
     postorder(root->rchild);
     printf("%d",root->data);
     num++;
     if(num<n) printf(" ");
};
int main()
{
     scanf("%d",&n);
     int val,index_p=,index_i=;
     char str[];
     stack<int> s;
     for(int i=;i<*n;i++)
     {
         scanf("%s",str);
         if(strcmp(str,"Push")==)
         {
             scanf("%d",&val);
             pre[index_p++]=val;
             s.push(val);
         }
         else
         {
             ino[index_i++]=s.top();
             s.pop();
         }
     }
     node *root=create(,n-,,n-);
     postorder(root);
     return ;
}