PAT 1020. Tree Traversals

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

思路就是递归，后续遍历根节点总是最后一个，然后在中序遍历中找到该值就可以判断左右数各有多少个，然后在递归，可以参考剑指Offer面试题：5.重建二叉树

import java.util.LinkedList;import java.util.Queue;import java.util.Scanner;public class Main {static class Node {Node left, right;int val;public Node(int val) {this.val = val;}}static int[] postOrder, inOrder;public static void main(String[] args) {Scanner sc = new Scanner(System.in);int N = sc.nextInt();postOrder = new int[N];inOrder   = new int[N];for(int i=0; i<N; i++)postOrder[i] = sc.nextInt();for(int i=0; i<N; i++)inOrder[i] = sc.nextInt();Node root = reconstruct(0, N-1, 0, N-1);levelOrder(root);}private static void levelOrder(Node root) {StringBuilder sb = new StringBuilder();Queue<Node> q = new LinkedList<Node>();q.add(root);while(!q.isEmpty()) {Node p = q.remove();sb.append(p.val + " ");if(p.left != null)q.add(p.left);if(p.right != null)q.add(p.right);}System.out.println(sb.toString().trim());}private static Node reconstruct(int ps, int pt, int is, int it) {if(ps == pt)return new Node(postOrder[ps]);int root_val = postOrder[pt];int root_position = 0;for(int i=is; i<=it; i++)if(inOrder[i] == root_val) {root_position = i;break;}Node root = new Node(root_val);//System.out.printf("%d %d %d %d\n", ps, ps+root_position-1-is, is, root_position-1);//System.out.printf("%d %d %d %d\n", ps+root_position-is, pt-1, root_position+1, it);if(ps+root_position-is-1 >= ps)root.left  = reconstruct(ps, ps+root_position-is-1, is, root_position-1);if(pt-1 >= ps+root_position-is)root.right = reconstruct(ps+root_position-is, pt-1, root_position+1, it);return root;}}