03-树3 Tree Traversals Again （25 分)

标签：char   and   each   rcm   roo   index   inpu   user   using   

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 (≤) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2 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

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
const int maxn = 35;
struct Node{
    int data;
    Node* lchild;
    Node* rchild;
};

int pre[maxn],in[maxn];
int n,num = 0;

Node* createTree(int preL,int preR,int inL,int inR){
    if(preL > preR) 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 numLeft = k - inL;
    root->lchild = createTree(preL+1,preL+numLeft,inL,k-1);
    root->rchild = createTree(preL+numLeft+1,preR,k+1,inR);
    return root;
}

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);
    char str[5];
    int x,preIndex = 0,inIndex = 0;
    stack<int> s;
    for(int i = 0; i < 2*n; i++){
        getchar();
        scanf("%s",str);
        if(strcmp(str,"Push") == 0){
            scanf("%d",&x);
            s.push(x);
            pre[preIndex++] = x;
        }else{
            in[inIndex++] = s.top();
            s.pop();
        }
    }
    Node* root = createTree(0,n-1,0,n-1);
    postOrder(root);
    return 0;
}

03-树3 Tree Traversals Again （25 分)

标签：char   and   each   rcm   roo   index   inpu   user   using   

原文地址：https://www.cnblogs.com/wanghao-boke/p/10669756.html