(PAT 1110) Complete Binary Tree (判断完全二叉树)

雨点打透心脏的1/2处 2022-04-23 05:54 141阅读 0赞

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
    - -
    - -

### Sample Output 1: ###

YES 8

### Sample Input 2: ###

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

### Sample Output 2: ###

NO 1

解题思路:

二叉树数组的存储方式是根据层序遍历来存储的

左子树的下标为root\*2 右子树下标为root\*2+1

我们通过先序遍历,每次遍历时带上结点的下标,遍历完成后,我们将得到一个最大下标

如果这个最大下标 == 结点个数,那么这颗树就是完全二叉树(因为我们知道,完全二叉树是被挤满的)

#include <iostream>
    #include <algorithm>
    #include <math.h>
    #include <string.h>
    #include <queue>
    using namespace std;
    const int MAXN = 50;
    struct MyStruct
    {
    	int lchild = -1;
    	int rchild = -1;
    	bool notRoot = false;
    }BTree[MAXN];
    int N;
    int maxn = -1;
    int ans = -1;
    void DFS(int root, int index) {
    	if (index > maxn) {
    		maxn = index;   //寻找最大下标
    		ans = root;
    	}
    	if (BTree[root].lchild != -1) DFS(BTree[root].lchild, index * 2);
    	if (BTree[root].rchild != -1) DFS(BTree[root].rchild, index * 2 + 1);
    }
    int main() {
    	scanf("%d", &N);
    	for (int i = 0; i < N; ++i) {
    		char lc[5], rc[5];
    		scanf("%*c%s %s", lc,rc);
    		if (strcmp(lc, "-") == 0 && strcmp(rc, "-") == 0) {
    			BTree[i].lchild = -1;
    			BTree[i].rchild = -1;
    			BTree[BTree[i].lchild].notRoot = true;
    			BTree[BTree[i].rchild].notRoot = true;
    
    		}
    		else if (strcmp(lc, "-") == 0 || strcmp(rc, "-") == 0) {
    			if (strcmp(lc, "-") == 0) {
    				BTree[i].lchild = -1;
    				BTree[i].rchild = atoi(rc);
    				BTree[BTree[i].rchild].notRoot = true;
    			}
    			else if (strcmp(rc, "-") == 0) {
    				BTree[i].lchild = atoi(lc);
    				BTree[i].rchild = -1;
    				BTree[BTree[i].lchild].notRoot = true;
    			}
    		}
    		else {
    			BTree[i].lchild = atoi(lc);
    			BTree[i].rchild = atoi(rc);
    			BTree[BTree[i].lchild].notRoot = true;
    			BTree[BTree[i].rchild].notRoot = true;
    		}
    	}
    	int nRoot = -1;
    	for (int i = 0; i < N; ++i) {
    		if (BTree[i].notRoot == false) {
    			nRoot = i;
    			break;
    		}
    	}
    	//遍历寻找最大下标
    	DFS(nRoot, 1);
    	if (maxn == N) {
    		cout << "YES" << " " << ans << endl;
    	}
    	else {
    		cout << "NO" << " " << nRoot << endl;
    	}
    
    	system("PAUSE");
    	return 0;
    }