顽强的小白

1086 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.

图片自己脑部

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

题目解析

用栈的形式来表示二叉树，这里就会发现一个有趣的现象，所有入栈的操作序列就是先序遍历，而所有出栈的序列就是中序遍历。发现这个现象后就和上一道二叉树的题目一样了，这次是给出中序和先序求后序遍历。

代码实现

#include <cstdio>
#include <algorithm>
#include <stack>
#include <cstring>
using namespace std;
const int maxn=35;
int n;
int in[maxn];
int pre[maxn];
int post[maxn];
struct node{int data;node* l;node* r;
};
node* create(int inL,int inR,int preL,int preR){if(inL>inR) return NULL;node* root=new node;root->data=pre[preL];int k;for(k=inL;k<=inR;++k){if(in[k]==pre[preL])break;}int left=k-inL;root->l=create(inL,k-1,preL+1,preL+left);root->r=create(k+1,inR,preL+left+1,preR);return root;
}
int cnt=0;
void postOrder(node* root){if(root==NULL) return;postOrder(root->l);postOrder(root->r);printf("%d",root->data);cnt++;if(cnt<n) printf(" ");}
int main(){stack<int> s;scanf("%d",&n);char mk[6];int j=0,k=0,x;for(int i=0;i<n*2;++i){scanf("%s",mk);if(strcmp(mk,"Push")==0){scanf("%d",&x);pre[j++]=x;s.push(x);}else{in[k++]=s.top();s.pop();}}node* root=create(0,n-1,0,n-1);postOrder(root);return 0;
}

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