1020. Tree Traversals (25)
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.
Input Specification:
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.
Output Specification:
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.
Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2
题目大意:给定一棵二叉树的后序遍历和中序遍历,请你输出其层序遍历的序列。这里假设键值都是互不相等的正整数。
PS:感谢github用户@fs19910227提供的pull request~
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 |
import java.util.LinkedList; import java.util.Scanner; public class Main { static Scanner scanner = new Scanner(System.in); static class Node { int value; Node left; Node right; Node(int value) { this.value = value; } @Override public String toString() { return "Node{" + "value=" + value + ", left=" + left + ", right=" + right + '}'; } } private static Node bulidTree() { int size = scanner.nextInt(); int[] postOrder = new int[size]; int[] inOrder = new int[size]; for (int i = 0; i < size; i++) { postOrder[i] = scanner.nextInt(); } for (int i = 0; i < size; i++) { inOrder[i] = scanner.nextInt(); } Node root = build(postOrder, inOrder, 0, size - 1, 0, size - 1); return root; } private static Node build(int[] postOrder, int[] inOrder, int postStart, int postEnd, int inStart, int inEnd) { if (postStart > postEnd) { return null; } if (postStart == postEnd) { return new Node(postOrder[postStart]); } int root = postOrder[postEnd--]; //find root in inOrder int inIndex = -1; for (int i = inStart; i <= inEnd; i++) { if (root == inOrder[i]) { inIndex = i; break; } } //recursion build int leftSize = inIndex - inStart; int rightSize = inEnd - inIndex; Node rootNode = new Node(root); rootNode.left = build(postOrder, inOrder, postStart, postStart + leftSize - 1, inStart, inIndex - 1); rootNode.right = build(postOrder, inOrder, postEnd - rightSize + 1, postEnd, inIndex + 1, inEnd); return rootNode; } public static void main(String[] args) { Node root = bulidTree(); LinkedList<Node> queue = new LinkedList<>(); queue.add(root); while (!queue.isEmpty()) { Node poll = queue.poll(); if (poll.left != null) { queue.add(poll.left); } if (poll.right != null) { queue.add(poll.right); } System.out.printf("%d%s", poll.value, queue.isEmpty() ? "\n" : " "); } } } |
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