138 lines
3.8 KiB
Java
138 lines
3.8 KiB
Java
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/** Author : Suraj Kumar
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* Github : https://github.com/skmodi649
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*/
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/** PROBLEM DESCRIPTION :
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* Given Preorder, Inorder and Postorder traversals of some tree of size N.
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* The task is to check if they are all of the same tree or not
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*/
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/** ALGORITHM :
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* The root element will be the first element of preorder.
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* Search for root in the inorder array and store it’s index as idx.
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* Use this idx to determine elements of left and right subtree in all three traversal arrays.
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* Call function recursively for both left and right sub tree.
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*/
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import java.lang.*;
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import java.util.*;
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class Tree{
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int val;
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Tree left, right;
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Tree(int val){
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this.val = val;
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left = null; right = null;
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}
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}
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class Main{
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static int idx;
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static boolean isPossible;
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public boolean checktree(int[] preorder, int[] inorder, int[] postorder, int N){
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idx = 0;
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isPossible = true;
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Map<Integer, Integer> hmap = new HashMap<>();
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for(int i = 0; i < inorder.length; i++) hmap.put(inorder[i], i);
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Tree root = buildTree(inorder, preorder, hmap, 0, N-1);
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if(!isPossible) return false;
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List<Integer> post = new ArrayList<>();
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buildPost(root, post);
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return Arrays.equals(post.stream().mapToInt(i->i).toArray(), postorder);
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}
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private static void buildPost(Tree root, List<Integer> post){
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if(root == null) return;
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buildPost(root.left, post);
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buildPost(root.right, post);
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post.add(root.val);
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}
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private static Tree buildTree(int[] inorder, int[] preorder, Map<Integer, Integer> hmap, int start, int end){
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if(start > end) return null;
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if(!isPossible) return null;
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int val = preorder[idx++];
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Tree root = new Tree(val);
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if(!hmap.containsKey(val)){
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isPossible = false;
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return null;
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}
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int pos = hmap.get(val);
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if(pos < start || pos > end){
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isPossible = false;
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return null;
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}
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root.left = buildTree(inorder, preorder, hmap, start, pos-1);
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root.right = buildTree(inorder, preorder, hmap, pos+1, end);
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return root;
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}
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public static void main(String args[])
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{
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Scanner sc = new Scanner(System.in);
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System.out.print("Enter the value of N : ");
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int n = sc.nextInt();
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int[] preorder = new int[n];
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int[] inorder = new int[n];
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int[] postorder = new int[n];
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System.out.println("Enter the elements of preorder array : ");
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for(int i=0; i<n; i++)
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preorder[i] = sc.nextInt();
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System.out.println("Enter the elements of inorder array : ");
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for(int i=0; i<n; i++)
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inorder[i] = sc.nextInt();
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System.out.println("Enter the elements of postorder array : ");
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for(int i=0; i<n; i++)
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postorder[i] = sc.nextInt();
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Main ob = new Main();
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if(ob.checktree(preorder, inorder, postorder, n) )
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System.out.println("Yes");
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else
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System.out.println("No");
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}
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}
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/** TEST CASES :
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* Test Case 1 :
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* Input : N = 5 , preorder[] = {1, 2, 4, 5, 3} , inorder[] = {4, 2, 5, 1, 3} and postorder[] = {4, 5, 2, 3, 1}
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* Output : Yes
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*
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* Test Case 2 :
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* Input : N = 5 , preorder[] = {1, 5, 4, 2, 3} , inorder[] = {4, 2, 5, 1, 3} and postorder[] = {4, 1, 2, 3, 5}
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* Output : No
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*/
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/** Explanation for Test Case 1 :
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* All the three traversals
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* are of the same tree.
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* 1
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* / \
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* 2 3
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* / \
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* 4 5
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* hence the output is Yes
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*/
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/** Time Complexity : O(N^2)
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* Auxiliary Space Complexity : O(N)
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* Constraints : 1 ≤ N ≤ 103
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* Node values are unique.
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*/
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