PATA1110题解

树的层序遍历判断完全二叉树

//
//  main.cpp
//  PATA1110
//
//  Created by Phoenix on 2018/2/23.
//  Copyright © 2018年 Phoenix. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
bool flag[21] = {false};
int n;
struct node {
    int data;
    int lchild, rchild;
}node[21];

bool flag1 = true, flag2 = true;
int lastnode, num = 0;
void level(int root) {
    queue<int> q;
    q.push(root);
    while (!q.empty()) {
        int top = q.front();
        //printf("%d ", top);
        num++;
        q.pop();
        if(top == -1 && flag1 == true) flag1 = false;
        if(flag1 == false && top >= 0) {
            flag2 = false;
            break;
        }
        if(num == n && flag1 == true) {
            lastnode = top;
            break;
        }
        if(top >= 0) {
            q.push(node[top].lchild);
            q.push(node[top].rchild);
        }
    }
}

int main(int argc, const char * argv[]) {
    scanf("%d", &n);
    for(int i = 0; i < n; i++) {
        char a[3], b[3];
        scanf("%s %s", a, b);
        if(a[0] == '-') node[i].lchild = -1;
        else {
            int ans = 0;
            for(int j = 0; j < strlen(a); j++) {
                ans = ans * 10 + a[j] - '0';
            }
            node[i].lchild = ans;
            flag[ans] = true;
        }
        if(b[0] == '-') node[i].rchild = -1;
        else {
            int ans = 0;
            for(int j = 0; j < strlen(b); j++) {
                ans = ans * 10 + b[j] - '0';
            }
            node[i].rchild = ans;
            flag[ans] = true;
        }
        //printf("%d %d\n", node[i].lchild, node[i].rchild);
    }
    int root;
    for(int i = 0; i < n; i++) {
        if(flag[i] == false) root = i;
    }
    level(root);
    if(flag2 == true) printf("YES %d\n", lastnode);
    else printf("NO %d\n", root);
    return 0;
}