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Commit 2ba8925

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Implement Binary Tree to Binary Search Tree Conversion - Solution
1 parent 43ed1bc commit 2ba8925

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module.exports = class Node {
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constructor(value) {
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this.value = value;
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this.leftChild = null; // will be a node
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this.rightChild = null; // will be a node
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}
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};
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const { binaryTreeToBST, storeInorder } = require('.');
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const Node = require('./Node');
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describe('Binary tree to binary search tree', () => {
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let tree;
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describe('Create Binary Tree', () => {
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tree = new Node(10);
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tree.leftChild = new Node(30);
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tree.leftChild.leftChild = new Node(20);
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tree.rightChild = new Node(15);
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tree.rightChild.rightChild = new Node(5);
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});
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it('Should converted binary tree to binary search tree', () => {
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binaryTreeToBST(tree);
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expect(storeInorder(tree)).toEqual([5, 10, 15, 20, 30]);
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});
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});
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/**
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* Given a Binary Tree, convert it to a Binary Search Tree.
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* The conversion must be done in such a way that keeps the original structure of Binary Tree.
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* Example 1
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Input:
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10
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/ \
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2 7
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/ \
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8 4
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Output:
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8
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/ \
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4 10
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/ \
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2 7
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*/
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// Helper function to store inorder traversal of a binary tree
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function storeInorder(root) {
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/** left - root - right */
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if (root === null) return [];
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// First store the left subtree
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let arr = [];
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const left = storeInorder(root.leftChild);
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arr = [...left, ...arr];
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// Append root's data
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arr = [...arr, root.value];
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// Store right subtree
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const right = storeInorder(root.rightChild);
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arr = [...arr, ...right];
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return arr;
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}
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// Helper function to copy elements from sorted array to make BST while keeping same structure
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function arrayToBST(arr, root) {
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const node = root;
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// Base case
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if (!node) return;
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// First update the left subtree
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arrayToBST(arr, node.leftChild);
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// update the root's data and remove it from sorted array
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node.value = arr.shift();
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// Finally update the right subtree
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arrayToBST(arr, node.rightChild);
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}
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function binaryTreeToBST(root) {
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// Tree is empty
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if (!root) return;
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const arr = storeInorder(root);
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arr.sort((a, b) => a - b);
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arrayToBST(arr, root);
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}
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module.exports = {
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binaryTreeToBST,
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storeInorder,
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};

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