#数据结构与算法学习笔记#PTA10：层次遍历叶节点（JAVA）

2018.4.9

要求寻找一棵二叉树的叶节点，并且按从上到下、从左到右的顺序打印出来。

题目难点在于如何层次遍历一颗二叉树，主要思想是利用队列层次遍历一棵树，先将根节点入队，每次出队时检查是否有左右子根节点，若有则子节点入队，直到队列为空。没有什么大坑。

Given a tree, you are supposed to list all the leaves in the order of top down, and left to right.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤10) 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 test case, print in one line all the leaves' indices in the order of top down, and left to right. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

class Node {
	int num;
	int leftChild;
	int rightChild;
}

public class List_Leaves {

	private static int num; // 结点数
	private static int root; // 根节点编号
	private static ArrayList<Node> tree = new ArrayList<Node>();
	private static ArrayList<Integer> leaves = new ArrayList<Integer>();
	private static Scanner indata = new Scanner(System.in);

	private static void Init() {

		num = Integer.parseInt(indata.nextLine());
		int[] check = new int[num]; // 检查根节点数组

		// 建树
		for (int i = 0; i < num; i++) {
			Node node = new Node();
			String[] data = indata.nextLine().split(" ");
			// 结点编号
			node.num = i;
			// 左子结点编号
			char left = data[0].charAt(0);
			if (left == '-') {
				node.leftChild = -1;
			} else {
				node.leftChild = left - '0';
				check[left - '0'] = 1;
			}
			// 右子节点编号
			char right = data[1].charAt(0);
			if (right == '-') {
				node.rightChild = -1;
			} else {
				node.rightChild = right - '0';
				check[right - '0'] = 1;
			}

			tree.add(node);
		}

		// 确定根节点
		for (int i = 0; i < num; i++) {
			if (check[i] == 0) {
				root = i;
				break;
			}
		}
	}
	
	private static void FindLeaves(){
		//利用队列层次遍历一棵树，先将根节点入队，每次出队时检查是否有左右子根节点，若有则子节点入队，直到队列为空
		Queue<Integer> queueTree = new LinkedList<Integer>();
		queueTree.add(root);
		
		while(!queueTree.isEmpty()){
			int node = queueTree.poll();
			//若左右子结点均为空，则为叶结点
			if(tree.get(node).leftChild == -1 && tree.get(node).rightChild == -1){
				leaves.add(node);
			}
			//若左子结点不空，则左子结点入栈
			if(tree.get(node).leftChild != -1){
				queueTree.add(tree.get(node).leftChild);
			}
			//若右子节点不空，则右子节点入栈
			if(tree.get(node).rightChild != -1){
				queueTree.add(tree.get(node).rightChild);
			}
		}

	}
	
	private static void Print(){
		if(!leaves.isEmpty()){
			System.out.print(leaves.get(0));
		}
		for(int i=1;i<leaves.size();i++){
			System.out.print(" " + leaves.get(i));
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Init();
		FindLeaves();
		Print();
	}

}