uva122 二叉树的层次遍历

ゞ浴缸里的玫瑰 2022-05-31 01:48 164阅读 0赞

Trees are fundamental in many branches of computer science (Pun definitely intended). Current stateof-the
    art parallel computers such as Thinking Machines’ CM-5 are based on fat trees. Quad- and
    octal-trees are fundamental to many algorithms in computer graphics.
    This problem involves building and traversing binary trees.
    Given a sequence of binary trees, you are to write a program
    that prints a level-order traversal of each tree. In this
    problem each node of a binary tree contains a positive integer
    and all binary trees have have fewer than 256 nodes.
    In a level-order traversal of a tree, the data in all nodes at
    a given level are printed in left-to-right order and all nodes at
    level k are printed before all nodes at level k + 1.
    For example, a level order traversal of the tree on the right
    is: 5, 4, 8, 11, 13, 4, 7, 2, 1.
    In this problem a binary tree is specified by a sequence
    of pairs ‘(n,s)’ where n is the value at the node whose path
    from the root is given by the string s. A path is given be
    a sequence of ‘L’s and ‘R’s where ‘L’ indicates a left branch and ‘R’ indicates a right branch. In the
    tree diagrammed above, the node containing 13 is specified by (13,RL), and the node containing 2 is
    specified by (2,LLR). The root node is specified by (5,) where the empty string indicates the path from
    the root to itself. A binary tree is considered to be completely specified if every node on all root-to-node
    paths in the tree is given a value exactly once.
    Input
    The input is a sequence of binary trees specified as described above. Each tree in a sequence consists
    of several pairs ‘(n,s)’ as described above separated by whitespace. The last entry in each tree is ‘()’.
    No whitespace appears between left and right parentheses.
    All nodes contain a positive integer. Every tree in the input will consist of at least one node and
    no more than 256 nodes. Input is terminated by end-of-file.
    Output
    For each completely specified binary tree in the input file, the level order traversal of that tree should
    be printed. If a tree is not completely specified, i.e., some node in the tree is NOT given a value or a
    node is given a value more than once, then the string ‘not complete’ should be printed.
    Sample Input
    (11,LL) (7,LLL) (8,R)
    (5,) (4,L) (13,RL) (2,LLR) (1,RRR) (4,RR) ()
    (3,L) (4,R) ()
    Sample Output
    5 4 8 11 13 4 7 2 1
    not complete

###### 题目大意：给你一个构造二叉树的规则，L代表构建左子树，R代表构建右子树，构建一颗二叉树，实现二叉树的层次遍历 ######

###### 解题思路：这是一个很简单的数据结构题目，首先通过给出的构造二叉树的规则，递归构建一颗二叉树，然后通过bfs用队列实现层次遍历，最后打印输出就可以了，把结果存在vecyor中即可， ######

###### 注意：在构建二叉树之前记得释放二叉树的存储空间，以防止内存泄漏remove\_tree为释放内存函数 ######

##### 代码： #####

#include <bits/stdc++.h>
    using namespace std;
    const int maxn = 1000;
    char s[maxn];
    struct Node
    {
        int v;
        bool have_value;
        Node *left,*right;
        Node():have_value(false),left(NULL),right(NULL){}
    };
    
    Node *root;
    bool failed;
    Node* newnode()
    {
        return new Node();
    }
    
    void remove_tree(Node *u)
    {
        if(u==NULL) return;
        remove_tree(u->left);
        remove_tree(u->right);
        delete u;
    }
    
    void addnode(int v,char *s)
    {
        int n = strlen(s);
        Node* u = root;
        for(int i=0;i<n;i++)
        {
            if(s[i]=='L')
            {
                if(u->left==NULL)
                    u->left = newnode();
                u = u->left;
            }
            else if(s[i]=='R')
            {
                if(u->right==NULL)
                    u->right = newnode();
                u = u->right;
            }
        }
        if(u->have_value) failed = true;
        u->v = v;
        u->have_value = true;
    }
    bool readin()
    {
        failed = false;
       for(;;)
        {
            if(scanf("%s",s)!=1)
               return false;
            if(!strcmp(s,"()"))
                break;
            int v;
            sscanf(&s[1],"%d",&v);
            addnode(v,strchr(s,',')+1);
        }
        return true;
    }
    
    bool bfs(vector<int>&ans)
    {
        queue<Node*>q;
        ans.clear();
        q.push(root);
        while(!q.empty())
        {
            Node *u = q.front();
            q.pop();
            if(!u->have_value)
                return false;
            ans.push_back(u->v);
            if(u->left!=NULL)
                q.push(u->left);
            if(u->right!=NULL)
                q.push(u->right);
        }
        return true;
    }
    
    int main()
    {
        //freopen("in.txt","r",stdin);
        while(1)
        {
            remove_tree(root);
            root =  newnode();
            if(!readin())
                break;
            vector<int>ans;
            if(!failed&&bfs(ans))
            {
                for(int i=0;i<ans.size();i++)
                    printf("%d%c",ans[i],i==ans.size()-1?'\n':' ');
            }
            else printf("not complete\n");
        }
        return 0;
    }