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  • PAT 1020 Tree Traversals

    1020 Tree Traversals (25 分)
     

    Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

    Input Specification:

    Each input file contains one test case. For each case, the first line gives a positive integer N (≤), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

    Output Specification:

    For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

    Sample Input:

    7
    2 3 1 5 7 6 4
    1 2 3 4 5 6 7
    

    Sample Output:

    4 1 6 3 5 7 2
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    
    struct node{
        int data = -1;
        node *left = NULL;
        node *right = NULL;
    };
    int n,cnt;
    
    int a[31];
    int b[31];
    
    node* createtree(int start,int end){
        if(start > end)
            return NULL;
        int num = a[cnt--];
        node *t = new(node);
        t->data = num;
        if(start == end){ return t;}
    
        int pos = -1;
        for(int i=start;i <= end;i++){
            if(b[i] == num) pos = i;
        }
    
        t->right = createtree(pos+1,end);
        t->left = createtree(start,pos-1);
    
        return t;
    }
    
    
    
    int main(){
        cin >> n;
        cnt = n-1;
    
        for(int i=0;i < n;i++) cin >> a[i];
        for(int i=0;i < n;i++) cin >> b[i];
    
        node *root = createtree(0,n-1);
    
        queue<node*> que;
        que.push(root);
        int count = 0;
    
        while(!que.empty()){
            node *t = que.front();que.pop();
            cout << t->data;
            if(count!=n-1){cout << " ";count++;}
            if(t->left)que.push(t->left);
            if(t->right)que.push(t->right);
        }
    
    
    
        return 0;
    }
    创建节点要new,不然就会出错
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  • 原文地址:https://www.cnblogs.com/cunyusup/p/10801579.html
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