pat1020. Tree Traversals (25)

算法思路：

1、后序最后元素为根，根将中序分为左右子树

2、层序遍历利用队列实现，java使用LinkedList

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

class Node{
	int key;
	Node left;
	Node right;
	public Node(int key){
		this.key=key;
	}
}
public class Main {
	static int num;
	static int post[];
	static int in[];
	public static Node createBT(int postH, int postT, int inH, int inT){
		if(postH>=postT)return null;
		int key=post[postT-1];
		Node root=new Node(key);
		int i=inH;
		for(;i<inT;i++)
			if(key==in[i])break;
		root.left=createBT(postH,i-inH+postH,inH,i);
		root.right=createBT(i-inH+postH,postT-1,i+1,inT);
		return root;
	}
	public static void levelTr(Node root){
		Queue<Node> que=new LinkedList<Node>();
		if(null!=root)System.out.print(root.key);
		if(null!=root.left)que.offer(root.left);
		if(null!=root.right)que.offer(root.right);
		while(!que.isEmpty()){
			Node node=que.poll();
			System.out.print(" "+node.key);
			if(null!=node.left)que.offer(node.left);
			if(null!=node.right)que.offer(node.right);
		}
	}
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		num=sc.nextInt();
		post=new int[num];
		in=new int[num];
		for(int i=0;i<num;i++)
			post[i]=sc.nextInt();
		for(int i=0;i<num;i++)
			in[i]=sc.nextInt();
		Node root=createBT(0,num,0,num);
		levelTr(root);
		sc.close();
	}
}

pat1020. Tree Traversals (25)

原文：http://blog.csdn.net/ldrmcml/article/details/44150997