• Complete Binary Tree


    Given a tree, you are supposed to tell if it is a complete binary tree.

    Input Specification:
    Each input file contains one test case. For each case, the first line gives a positive integer N (≤20) which is the total number of nodes in the tree – and hence the nodes are numbered from 0 to N−1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a - will be put at the position. Any pair of children are separated by a space.

    Output Specification:
    For each case, print in one line YES and the index of the last node if the tree is a complete binary tree, or NO and the index of the root if not. There must be exactly one space separating the word and the number.

    Sample Input 1:

    9
    7 8
    - -
    - -
    - -
    0 1
    2 3
    4 5
    - -
    - -
    
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    Sample Output 1:

    YES 8
    
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    Sample Input 2:

    8
    - -
    4 5
    0 6
    - -
    2 3
    - 7
    - -
    - -
    
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    Sample Output 2:

    NO 1
    
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    题目大意:
    判断一棵树是不是完全二叉树,是的话输出YES和最后一个结点,否则输出NO和根节点。

    王道书上同样有相同的一道算法题,算法的思路都一样,只是王道书上要求的树是用二叉链表存储结构存储的。

    思路:
    采用层次遍历算法,将所有结点加入队列(包括空结点)。遇到空结点时,看其后是否有非空结点。若有,则二叉树不是完全二叉树。

    #include
    using namespace std;
    struct Complete_Binary_Tree
    {
        int data;
        int lchild, rchild;
    }node[25];
    int n;
    bool visited[25] = {false};
    int turn(string s) {
        int sum = 0;
        for(auto &i:s) sum = sum * 10 + (i - '0');
        return sum;
    }
    int build_tree() {
        scanf("%d", &n);
        for(int i = 0; i < n; i++) {
            string left, right;
            cin >> left >> right;
            node[i].data = i;
            if(left == "-") node[i].lchild = -1;
            else {
                node[i].lchild = turn(left);
                visited[node[i].lchild] = true;
            }
            if(right == "-") node[i].rchild = -1;
            else {
                node[i].rchild = turn(right);
                visited[node[i].rchild] = true;
            }
        }
        for(int i = 0; i < n; i++) {
            if(!visited[i]) return i;//返回根节点
        }
    }
    int last_node = 0;
    bool iscomplete(int root) {
        queue<int>q;
        if(n == 0) return true; // 空树为满二叉树
        q.push(root);
        while (!q.empty())
        {
            int top = q.front();
            q.pop();
            if(top != -1) last_node = top;
            if(top != -1) {
                q.push(node[top].lchild);
                q.push(node[top].rchild);
            } else { // 结点为空,检查其后是否有非空结点
                while(!q.empty()) {
                    int temp = q.front();
                    q.pop();
                    if(temp != -1) return false; // 结点非空,则二叉树为非完全二叉树
                }
            }
        }
        return true;
    }
    int main() {
        int root = build_tree();
        // for(int i = 0; i < n; i++) cout << node[i].lchild << " " << node[i].rchild << endl;
        // cout << root << endl;
        if(iscomplete(root)) printf("YES %d", last_node);
        else printf("NO %d", root);
        return 0;
    }
    
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  • 原文地址:https://blog.csdn.net/liupang14159/article/details/125911674