PAT1086:Tree Traversals Again【Java实现】——改进版

之前的解法是：得到前序和中序遍历，建立二叉树，然后输出后序遍历

后来经过陈越老师的讲解才知道，其实没有必要去建立二叉树，知道前序和中序的情况下，可以直接利用递归求得后续遍历

Java代码如下：

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

/**
 * Tree Traversals Again 之前已经写了一篇博客，但后来发现那并不是最优的解法
 * 原来知道树的前序中序遍历以后并不需要建树，可以通过递归直接写出后续遍历
 * 
 * @author Jacob
 *
 */

public class TreeTraversalsAgain {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int NODE_NUM = sc.nextInt();
		int[] preOrder = new int[NODE_NUM];
		int[] inOrder = new int[NODE_NUM];
		int preCount = 0;
		int inCount = 0;
		Stack stack = new Stack();
		int tmp;
		String operate;// 用来保存push或者pop

		// 通过堆栈操作序列得到前序遍历和中序遍历
		// 注意：字符串比较不能用“==”，要用equals()
		for (int i = 0; i < NODE_NUM * 2; i++) {
			operate = sc.next();
			if (operate.equals("Push")) {
				tmp = sc.nextInt();
				stack.push(tmp);
				preOrder[preCount] = tmp;
				preCount++;
			} else if (operate.equals("Pop")) {
				inOrder[inCount] = stack.pop();
				inCount++;
			}
		}
		//
		int[] postOrder = new int[NODE_NUM];
		solve(preOrder, inOrder, postOrder, 0, 0, 0, NODE_NUM);

		for (int i = 0; i < NODE_NUM; i++) {
			System.out.print(postOrder[i]);
			if(i!=NODE_NUM-1){
				System.out.print(" ");
			}
		}

		sc.close();
	}

	public static void solve(int[] preOrder, int[] inOrder, int[] postOrder, int preLeft, int inLeft, int postLeft,
			int num) {
		if (num == 0) {
		}
		if (num == 1)
			postOrder[postLeft] = preOrder[preLeft];
		if (num > 1) {
			int findNum = 0; // 存储中间节点的下标
			for (int i = inLeft; i < inLeft + num; i++) {
				if (preOrder[preLeft] == inOrder[i]) {
					findNum = i;
					break;
				}
			}
			postOrder[postLeft + num - 1] = preOrder[preLeft];
			int leftNum = findNum - inLeft;
			int rightNum = num - leftNum - 1;
			solve(preOrder, inOrder, postOrder, preLeft + 1, inLeft, postLeft, leftNum);
			solve(preOrder, inOrder, postOrder, preLeft + leftNum + 1, inLeft + leftNum + 1, postLeft + leftNum,
					rightNum);
		}
	}
}

/**
 * 树节点类
 * 
 */
class Node {
	int value;
	Node left;
	Node right;

	Node(int value) {
		this.value = value;
		left = null;
		right = null;
	}
}