Merge pull request knaxus#127 from k88manish/master

Implement Binary Tree to Binary Search Tree Conversion - Solution

Author: TheSTL

src/_DataStructures_/Trees/BinaryTree/index.js

@@ -3,14 +3,14 @@ const Node = require('./Node');
 class BinaryTree {
   constructor(arr) {
     if (!Array.isArray(arr) || !arr.length) {
-      throw new Error('Invalid argument to create a Binary Tre');
+      throw new Error('Invalid argument to create a Binary Tree');
     }
     this.root = this.createBinaryTree((this.root = null), arr, 0);
   }
 
   // eslint-disable-next-line class-methods-use-this
   createBinaryTree(root, arr, i) {
-    if (i < arr.length) {
+    if (i < arr.length && arr[i]) {
       // eslint-disable-next-line no-param-reassign
       root = new Node(arr[i]);
       // eslint-disable-next-line no-param-reassign

src/_Problems_/binary-tree-to-binary-search-tree/binary-tree-to-binary-search-tree.test.js

@@ -0,0 +1,15 @@
+const { binaryTreeToBST, storeInorder } = require('.');
+const BinaryTree = require('../../_DataStructures_/Trees/BinaryTree');
+
+describe('Binary tree to binary search tree', () => {
+  let tree;
+
+  describe('Create Binary Tree', () => {
+    tree = new BinaryTree([10, 30, 15, 20, null, null, 5]);
+  });
+
+  it('Should converted binary tree to binary search tree', () => {
+    const bTree = binaryTreeToBST(tree);
+    expect(storeInorder(bTree)).toEqual([5, 10, 15, 20, 30]);
+  });
+});

src/_Problems_/binary-tree-to-binary-search-tree/index.js

@@ -0,0 +1,78 @@
+/**
+ * Given a Binary Tree, convert it to a Binary Search Tree.
+ * The conversion must be done in such a way that keeps the original structure of Binary Tree.
+ * Example 1
+Input:
+          10
+         /  \
+        2    7
+       / \
+      8   4
+Output:
+          8
+         /  \
+        4    10
+       / \
+      2   7
+ */
+
+const Node = require('../../_DataStructures_/Trees/BinaryTree/Node');
+// Helper function to store inorder traversal of a binary tree
+function storeInorder(root) {
+  /** left - root - right */
+  if (root === null) return [];
+
+  // First store the left subtree
+  let arr = [];
+  const left = storeInorder(root.leftChild);
+  arr = [...left, ...arr];
+
+  // Append root's data
+  arr = [...arr, root.value];
+
+  // Store right subtree
+  const right = storeInorder(root.rightChild);
+  arr = [...arr, ...right];
+  return arr;
+}
+
+// Helper function to copy elements from sorted array to make BST while keeping same structure
+// Runtime complexity iof this function is O(n) where n is number of nodes, as we are each node of tree one time.
+function arrayToBST(arr, root) {
+  const node = root;
+  // Base case
+  if (!node) return null;
+
+  const bstNode = new Node();
+  // First update the left subtree
+  const leftChild = arrayToBST(arr, node.leftChild);
+  if (leftChild) {
+    bstNode.leftChild = leftChild;
+  }
+
+  // update the root's data and remove it from sorted array
+  // eslint-disable-next-line no-param-reassign
+  bstNode.value = arr.shift();
+
+  // Finally update the right subtree
+  const rightChild = arrayToBST(arr, node.rightChild);
+  if (rightChild) {
+    bstNode.rightChild = rightChild;
+  }
+
+  return bstNode;
+}
+
+function binaryTreeToBST(bTree) {
+  // Tree is empty
+  if (!bTree.root) return null;
+  const arr = bTree.preOrder();
+  arr.sort((a, b) => a - b);
+  const bst = arrayToBST(arr, bTree.root);
+  return bst;
+}
+
+module.exports = {
+  binaryTreeToBST,
+  storeInorder,
+};