1110 Complete Binary Tree (25分)

本文介绍了一种判断给定树是否为完全二叉树的算法,并提供了详细的输入输出规格及实现代码。通过层次遍历的方式,检查树的每个节点及其子节点,以确定其是否满足完全二叉树的条件。

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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.

A1123结合起来看

#include<iostream>
#include<queue>
using namespace std;
struct tree {
	int l = -1, r = -1;
};
tree t[30];
int mark[30] = { 0 }, n, root, index;
bool level(int root,int &index) {
	bool isCBT = true, flag = false;
	queue<int>q;
	q.push(root);
	int cnt = 0;
	while (!q.empty()) {
		int p = q.front();
		q.pop();
		if (p != -1) {
			index = p;
			if (flag)isCBT = false;
			q.push(t[p].l);
			q.push(t[p].r);
		}
		else
			flag = true;
	}
	return isCBT;
}
int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		string l, r;
		cin >> l >> r;
		if (l != "-") {
			t[i].l = atoi(l.c_str());
			mark[atoi(l.c_str())] = 1;
		}
		if (r != "-") {
			t[i].r = atoi(r.c_str());
			mark[atoi(r.c_str())] = 1;
		}
	}
	for (int i = 0; i < n; i++) {
		if (mark[i] == 0) {
			root = i;
			break;
		}
	}
	if (level(root, index))cout << "YES " << index << endl;
	else
		cout << "NO " << root << endl;
	return 0;
}
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