freeCodeCamp/curriculum/challenges/english/08-coding-interview-prep/data-structures/invert-a-binary-tree.english.md

id	title	challengeType
587d8259367417b2b2512c83	Invert a Binary Tree	1

Description

Here will we create a function to invert a binary tree. Given a binary tree, we want to produce a new tree that is equivalently the mirror image of this tree. Running an inorder traversal on an inverted tree will explore the nodes in reverse order when compared to the inorder traversal of the original tree. Write a method to do this called invert on our binary tree. Calling this method should invert the current tree structure. Ideally, we would like to do this in-place in linear time. That is, we only visit each node once and we modify the existing tree structure as we go, without using any additional memory. Good luck!

Instructions

Tests

tests:
  - text: The <code>BinarySearchTree</code> data structure exists.
    testString: assert((function() { var test = false; if (typeof BinarySearchTree !== 'undefined') { test = new BinarySearchTree() }; return (typeof test == 'object')})(), 'The <code>BinarySearchTree</code> data structure exists.');
  - text: The binary search tree has a method called <code>invert</code>.
    testString: assert((function() { var test = false; if (typeof BinarySearchTree !== 'undefined') { test = new BinarySearchTree() } else { return false; }; return (typeof test.invert == 'function')})(), 'The binary search tree has a method called <code>invert</code>.');
  - text: The <code>invert</code> method correctly inverts the tree structure.
    testString: assert((function() { var test = false; if (typeof BinarySearchTree !== 'undefined') { test = new BinarySearchTree() } else { return false; }; if (typeof test.invert !== 'function') { return false; }; test.add(4); test.add(1); test.add(7); test.add(87); test.add(34); test.add(45); test.add(73); test.add(8); test.invert(); return test.inorder().join('') == '877345348741'; })(), 'The <code>invert</code> method correctly inverts the tree structure.');
  - text: Inverting an empty tree returns <code>null</code>.
    testString: assert((function() { var test = false; if (typeof BinarySearchTree !== 'undefined') { test = new BinarySearchTree() } else { return false; }; if (typeof test.invert !== 'function') { return false; }; return (test.invert() == null); })(), 'Inverting an empty tree returns <code>null</code>.');

Challenge Seed

var displayTree = (tree) => console.log(JSON.stringify(tree, null, 2));
function Node(value) {
    this.value = value;
    this.left = null;
    this.right = null;
}
function BinarySearchTree() {
    this.root = null;
    // change code below this line
    // change code above this line
}

After Test

BinarySearchTree.prototype = {
    add: function(value) {
        var node = this.root;
        if (node == null) {
          this.root = new Node(value);
          return;
        } else {
            function searchTree(node) {
                if (value < node.value) {
                    if (node.left == null) {
                        node.left = new Node(value);
                        return;
                    } else if (node.left != null) {
                        return searchTree(node.left)
                    };
                } else if (value > node.value) {
                    if (node.right == null) {
                        node.right = new Node(value);
                        return;
                    } else if (node.right != null) {
                        return searchTree(node.right);
                    };
                } else {
                    return null;
                };
            };
            return searchTree(node);
        };
    },
    inorder: function() {
        if (this.root == null) {
          return null;
        } else {
          var result = new Array();
          function traverseInOrder(node) {
              if (node.left != null) {
                  traverseInOrder(node.left);
              };
              result.push(node.value);
              if (node.right != null) {
                  traverseInOrder(node.right);
              };
          }
          traverseInOrder(this.root);
          return result;
        };
    }
};

Solution

// solution required