1110 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 (≤) 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
- -
- -

 

Sample Output 1:

YES 8

 

Sample Input 2:

8
- -
4 5
0 6
- -
2 3
- 7
- -
- -

 

Sample Output 2:

NO 1

题意：

　　给出一棵树中各个结点的左右孩子的信息，判断这棵树是不是完全二叉树，如果是，输出最后一个结点的下标，如果不是，输出根节点的下标。

思路：

　　刚开始想的是把这棵树给构建出来，然后按照要求输出就好了，但是，这样做有三组测试样例不能通过。（找到错误的原因了，是我把字符串转数字的函数写错了，其实可以不用写的，直接用stoi()函数就好了）然后，看了一下别人的代码，发现可以不用构建树，就可以做出来。（既然我的代码已经通过了，就把我的写上去吧，哈哈哈）

Code：

#include <bits/stdc++.h>

using namespace std;

map<int, pair<int, int> > m;

typedef struct Node* node;

struct Node {
    int val;
    node left;
    node right;
    Node(int v) {
        val = v;
        left = NULL;
        right = NULL;
    }
};

int lastNode(node root) {
    queue<node> que;
    que.push(root);
    bool haveEnd = false;
    node temp;
    while (!que.empty()) {
        temp = que.front();
        que.pop();
        if (temp->left != NULL) {
            if (haveEnd) return -1;
            que.push(temp->left);
        } else
            haveEnd = true;
        if (temp->right != NULL) {
            if (haveEnd) return -1;
            que.push(temp->right);
        } else
            haveEnd = true;
    }
    return temp->val;
}

node buildTree(int r) {
    if (r < 0) return NULL;
    node root = new Node(r);
    if (m.find(r) != m.end() && m[r].first != -1) 
        root->left = buildTree(m[r].first);
    if (m.find(r) != m.end() && m[r].second != -1) 
        root->right = buildTree(m[r].second);
    return root;
}

int main() {
    int n;
    cin >> n;
    vector<bool> haveParent(n, false);
    for (int i = 0; i < n; ++i) {
        string left, right;
        cin >> left >> right;
        pair<int, int> temp = {-1, -1};
        if (left[0] != '-') {
            int l = stoi(left);
            haveParent[l] = true;
            temp.first = l;
        }
        if (right[0] != '-') {
            int r = stoi(right);
            haveParent[r] = true;
            temp.second = r;
        }
        m[i] = temp;
    }

    int r = 0;
    while(haveParent[r]) r++;

    node root = buildTree(r);
    int last = lastNode(root);
    if (last < 0)
        cout << "NO " << root->val << endl;
    else
        cout << "YES " << last << endl;

    return 0;
}