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| 1 | +package DataStructures.Trees; |
| 2 | + |
| 3 | +import java.util.HashMap; |
| 4 | +import java.util.Map; |
| 5 | +import DataStructures.Trees.BinaryTree.Node; |
| 6 | + |
| 7 | +/** |
| 8 | + * Approach: Naive Solution: Create root node from first value present in |
| 9 | + * preorder traversal. Look for the index of root node's value in inorder |
| 10 | + * traversal. That will tell total nodes present in left subtree and right |
| 11 | + * subtree. Based on that index create left and right subtree. |
| 12 | + * Complexity: |
| 13 | + * Time: O(n^2) for each node there is iteration to find index in inorder array |
| 14 | + * Space: Stack size = O(height) = O(lg(n)) |
| 15 | + * |
| 16 | + * Optimized Solution: Instead of iterating over inorder array to find index of |
| 17 | + * root value, create a hashmap and find out the index of root value. |
| 18 | + * Complexity: |
| 19 | + * Time: O(n) hashmap reduced iteration to find index in inorder array |
| 20 | + * Space: O(n) space taken by hashmap |
| 21 | + * |
| 22 | + */ |
| 23 | +public class CreateBinaryTreeFromInorderPreorder { |
| 24 | + public static void main(String[] args) { |
| 25 | + test(new Integer[] {}, new Integer[] {}); // empty tree |
| 26 | + test(new Integer[] { 1 }, new Integer[] { 1 }); // single node tree |
| 27 | + test(new Integer[] { 1, 2, 3, 4 }, new Integer[] { 1, 2, 3, 4 }); // right skewed tree |
| 28 | + test(new Integer[] { 1, 2, 3, 4 }, new Integer[] { 4, 3, 2, 1 }); // left skewed tree |
| 29 | + test(new Integer[] { 3, 9, 20, 15, 7 }, new Integer[] { 9, 3, 15, 20, 7 }); // normal tree |
| 30 | + } |
| 31 | + |
| 32 | + private static void test(final Integer[] preorder, final Integer[] inorder) { |
| 33 | + System.out.println("\n===================================================="); |
| 34 | + System.out.println("Naive Solution..."); |
| 35 | + BinaryTree root = new BinaryTree(createTree(preorder, inorder, 0, 0, inorder.length)); |
| 36 | + System.out.println("Preorder Traversal: "); |
| 37 | + root.preOrder(root.getRoot()); |
| 38 | + System.out.println("\nInorder Traversal: "); |
| 39 | + root.inOrder(root.getRoot()); |
| 40 | + System.out.println("\nPostOrder Traversal: "); |
| 41 | + root.postOrder(root.getRoot()); |
| 42 | + |
| 43 | + Map<Integer, Integer> map = new HashMap<>(); |
| 44 | + for (int i = 0; i < inorder.length; i++) { |
| 45 | + map.put(inorder[i], i); |
| 46 | + } |
| 47 | + BinaryTree optimizedRoot = new BinaryTree(createTreeOptimized(preorder, inorder, 0, 0, inorder.length, map)); |
| 48 | + System.out.println("\n\nOptimized solution..."); |
| 49 | + System.out.println("Preorder Traversal: "); |
| 50 | + optimizedRoot.preOrder(root.getRoot()); |
| 51 | + System.out.println("\nInorder Traversal: "); |
| 52 | + optimizedRoot.inOrder(root.getRoot()); |
| 53 | + System.out.println("\nPostOrder Traversal: "); |
| 54 | + optimizedRoot.postOrder(root.getRoot()); |
| 55 | + } |
| 56 | + |
| 57 | + private static Node createTree(final Integer[] preorder, final Integer[] inorder, |
| 58 | + final int preStart, final int inStart, final int size) { |
| 59 | + if (size == 0) { |
| 60 | + return null; |
| 61 | + } |
| 62 | + |
| 63 | + Node root = new Node(preorder[preStart]); |
| 64 | + int i = inStart; |
| 65 | + while (preorder[preStart] != inorder[i]) { |
| 66 | + i++; |
| 67 | + } |
| 68 | + int leftNodesCount = i - inStart; |
| 69 | + int rightNodesCount = size - leftNodesCount - 1; |
| 70 | + root.left = createTree(preorder, inorder, preStart + 1, inStart, leftNodesCount); |
| 71 | + root.right = createTree(preorder, inorder, preStart + leftNodesCount + 1, i + 1, |
| 72 | + rightNodesCount); |
| 73 | + return root; |
| 74 | + |
| 75 | + } |
| 76 | + |
| 77 | + private static Node createTreeOptimized(final Integer[] preorder, final Integer[] inorder, |
| 78 | + final int preStart, final int inStart, final int size, |
| 79 | + final Map<Integer, Integer> inorderMap) { |
| 80 | + if (size == 0) { |
| 81 | + return null; |
| 82 | + } |
| 83 | + |
| 84 | + Node root = new Node(preorder[preStart]); |
| 85 | + int i = inorderMap.get(preorder[preStart]); |
| 86 | + int leftNodesCount = i - inStart; |
| 87 | + int rightNodesCount = size - leftNodesCount - 1; |
| 88 | + root.left = createTreeOptimized(preorder, inorder, preStart + 1, inStart, |
| 89 | + leftNodesCount, inorderMap); |
| 90 | + root.right = createTreeOptimized(preorder, inorder, preStart + leftNodesCount + 1, |
| 91 | + i + 1, rightNodesCount, inorderMap); |
| 92 | + return root; |
| 93 | + } |
| 94 | + |
| 95 | +} |
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