Tree Traversals Again（java实现）

7-5 Tree Traversals Again（25 分）

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

 

Sample Input:

6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop

Sample Output:

3 4 2 6 5 1

思路：

1、主要是根据中序遍历堆栈push和pop的操作来建立树的过程。

2、走一遍就会发现一个规律，对于push操作而言，如果它的前一个操作是pop，那么push进来的这个节点就是前一个pop出来的节点的右孩子。如果它的前一个操作是push，那这个节点就是前一个push进来的节点的左孩子。用循环结合stack容器即可把树用静态链表建立起来。

3、树建好之后，用递归即可把树用后序遍历打印出来。

4、几个单词记住。inorder 中序；preorder 前序； postorder 后序； binary tree 二叉树；traversal 遍历；

 

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看了标准答案，是直接根据前序和中序的数组递归求出后序遍历的数组。还是建议用堆栈的方法模拟系统的递归实现。

import java.util.Scanner;
import java.util.Stack;

public class Main {
	public static int flag=0;

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);	
		Jiedian [] tree = build(in);
//		for(Jiedian jd :tree) {
//			System.out.println(jd.element+" "+jd.left+" "+jd.right);
//		}
		int root=0;
		print(tree,root);
		
		
	}

	private static void print(Jiedian[] tree,int root) {              //用递归后序遍历输出
		// TODO Auto-generated method stub
	
		if(root!=-1)
		{
			print(tree,tree[root].left);
			print(tree,tree[root].right);
			if(flag==0)
			{
				System.out.print(tree[root].element);
				flag++;
			}
			else
				System.out.print(" "+tree[root].element);
		}

	}

	private static Jiedian[] build(Scanner in) {
		// TODO Auto-generated method stub
		int n = in.nextInt();
		Jiedian [] jd = new Jiedian[n];
		String [] str = new String[2*n];
		Stack<Jiedian> st = new Stack<>();
		
		
		for(int i=0,j=0,k=0;i<2*n;i++) {						//读入数据，建立push和pop操作的String数组
			str[i]=new String(in.next());				
			Jiedian temp = new Jiedian();						//建立按push顺序建立的Jiedian数组。
			if(str[i].equals("Push")) {
				temp.element=in.nextInt();
				jd[j]=temp;
				j++;
			}
		}

		st.push(jd[0]);
		
		Jiedian temp = new Jiedian();
		Jiedian pop = new Jiedian();
		
		for(int i =1,j=1;i<2*n;i++) {                                    //用静态链表建立树
			if(str[i].equals("Push")) {
				if(str[i-1].equals("Push")) {
						temp=st.peek();
						temp.left=j;
						st.push(jd[j]);
						j++;
						
				}
				else
					{
						pop.right=j;
						st.push(jd[j]);
						j++;
					}
			}
			else
				pop=st.pop();
		}
		return jd;
	}
}
class Jiedian{
	public int local;
	public int element;
	public int left=-1;
	public int right=-1;
}