PTA——03-树3 Tree Traversals Again（25 分）【java语言实现】

03-树3 Tree Traversals Again（25 分）

题目

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） 通过 输入的 push 顺序，得到前序遍历的数组

2）通过 push 和 pop 得到 中序遍历的数组

3）通过前序和中序的数组，可以唯一确定一个后序遍历数组。

代码

import java.util.*;

class TreeNode1{
    int val;
    int left;
    int right;
}

/**
 * 拥有前序和中序，需要后序遍历结点
 */
public class Main {

//    public static int[] in ={};
//    int[] pre ={};

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N =sc.nextInt();
      //  int root =buildTree(sc,N,T1);

        //list 中第一个为前序，第二个为中序，需要打印的是后序遍历的结果
        List> list = buildTree(sc,N);

        list.get(0);
        int[] pre = new int[list.get(0).size()];
        int[] in  = new int[list.get(1).size()];

        for (int i=0;i0).get(i);
        }
        for (int i=0;i1).get(i);
        }

        List res = new ArrayList<>();

        post(0,0,in.length-1,in,pre,res);
        //System.out.println();
        for (int i=0;iif (i!= res.size()-1){
                System.out.print(res.get(i)+" ");
            }
            else
                System.out.print(res.get(i));
        }




        //知道前序中序，需要打印后序
        //前序：根左右；中序：左根右

        System.out.println();
    }

    private static List> buildTree(Scanner sc, int N){
        Stack stack = new Stack<>();//构建一个堆
        List preOrderTrav = new ArrayList<>();
        List middleOrder = new ArrayList<>();

        for (int i=0;i<2*N;i++){
            String str = sc.next();
            if (str.equals("Push")){
                int num = sc.nextInt();
                stack.push(num);
                preOrderTrav.add(num);
            }
            else {
                int num = stack.pop();
                middleOrder.add(num);
            }
        }

        List> res = new ArrayList<>();
        res.add(preOrderTrav);
        res.add(middleOrder);

        return res;
    }


    /**
     * https://blog.csdn.net/liuchuo/article/details/52135882
     * @param root
     * @param start
     * @param end
     * @param in
     * @param pre
     * @param res
     * @return
     */
   public static List post(int root, int start, int end,int [] in,int []pre,List res) {
        if(start > end)
            return null;
        int i = start;
        while(i < end && in[i] != pre[root]) i++;
        post(root + 1, start, i - 1,in,pre,res);
        post(root + 1 + i - start, i + 1, end,in,pre,res);
        res.add(pre[root]);
        //System.out.println(pre[root]);

        return res;
    }


}

转载于:https://www.cnblogs.com/HuanChen1025/p/8999275.html