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~
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" : " ");
}
}
}