标签：node 1086 PAT Again int tree preL pop root

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.

#include <cstdio>
#include <cstring>
#include <stack>
//已知树的前序和中序，输出树的后序
using namespace std;
struct node{
    int data;
    node* lchild;
    node* rchild;
};
int pre[35] , in[35] ;//先序，中序
int N;//节点个数
stack<int> nums;
//建树
node* creat(int preL, int preR, int inL, int inR)
{
    if(preL > preR) return nullptr;
    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 = creat(preL + 1, preL + numLeft,inL, k-1);
    root->rchild = creat(preL + numLeft + 1, preR, k +1, inR);
    return root;
}
//后序输出
int flag = 0;
void postorder(node* root)
{

    if(root == nullptr) return;
    postorder(root->lchild);
    postorder(root->rchild);
    printf("%d", root->data);
    flag++;
    if(flag < N)printf(" ");
}

int main()
{
    char str[10];
    int num, k = 0,n = 0;
    scanf("%d", &N);
    for (int i = 0; i<2*N; ++i)
    {
        scanf("%s",&str);
        if(!strcmp(str, "Push"))
        {
            scanf("%d",&num);
            nums.push(num);
            pre[k] = num;
            k++;
        } else
        {
            in[n] = nums.top();
            nums.pop();
            n++;
        }
    }
    node* root = creat(0, N-1, 0,N-1);
    postorder(root);
    return 0;
}

标签：node,1086,PAT,Again,int,tree,preL,pop,root
来源： https://www.cnblogs.com/xinyuLee404/p/12626912.html

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