Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.
Each input file contains one test case. For each case, the first line gives a positive integer N (≤30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
4 1 6 3 5 7 2
1 package pattest; 2 3 import java.util.LinkedList; 4 import java.util.Scanner; 5 6 /** 7 * @Auther: Xingzheng Wang 8 * @Date: 2019/2/28 15:52 9 * @Description: pattest 10 * @Version: 1.0 11 */ 12 public class PAT1020 { 13 static Scanner scanner = new Scanner(System.in); 14 15 static class Node { 16 int value; 17 Node left; 18 Node right; 19 20 Node(int value) { 21 this.value = value; 22 } 23 24 @Override 25 public String toString() { 26 return "Node{" + "value=" + value + ", left=" + left + ", right=" + right + ‘}‘; 27 } 28 } 29 30 private static Node bulidTree() { 31 int size = scanner.nextInt(); 32 int[] postOrder = new int[size]; 33 int[] inOrder = new int[size]; 34 for (int i = 0; i < size; i++) { 35 postOrder[i] = scanner.nextInt(); 36 } 37 for (int i = 0; i < size; i++) { 38 inOrder[i] = scanner.nextInt(); 39 } 40 Node root = build(postOrder, inOrder, 0, size - 1, 0, size - 1); 41 return root; 42 } 43 44 private static Node build(int[] postOrder, int[] inOrder, int postStart, int postEnd, int inStart, int inEnd) { 45 if (postStart > postEnd) { 46 return null; 47 } 48 if (postStart == postEnd) { 49 return new Node(postOrder[postStart]); 50 } 51 int root = postOrder[postEnd--]; 52 53 //find root in inOrder 54 int inIndex = -1; 55 for (int i = inStart; i <= inEnd; i++) { 56 if (root == inOrder[i]) { 57 inIndex = i; 58 break; 59 } 60 } 61 //recursion build 62 int leftSize = inIndex - inStart; 63 int rightSize = inEnd - inIndex; 64 Node rootNode = new Node(root); 65 rootNode.left = build(postOrder, inOrder, postStart, postStart + leftSize - 1, inStart, inIndex - 1); 66 rootNode.right = build(postOrder, inOrder, postEnd - rightSize + 1, postEnd, inIndex + 1, inEnd); 67 return rootNode; 68 } 69 70 public static void main(String[] args) { 71 Node root = bulidTree(); 72 LinkedList<Node> queue = new LinkedList<>(); 73 queue.add(root); 74 while (!queue.isEmpty()) { 75 Node poll = queue.poll(); 76 if (poll.left != null) { 77 queue.add(poll.left); 78 } 79 if (poll.right != null) { 80 queue.add(poll.right); 81 } 82 System.out.printf("%d%s", poll.value, queue.isEmpty() ? "\n" : " "); 83 } 84 } 85 }
[PAT] 1020 Tree Traversals （25 分）Java
原文：https://www.cnblogs.com/PureJava/p/10498018.html