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Stan Choi a13d116aab fix(learn): Added solution for invert binary tree data structures challenge (#35922)
* fix: add solution for invert binary tree

* Fixed errors

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* Update guide/english/certifications/coding-interview-prep/data-structures/invert-a-binary-tree/index.md

Co-Authored-By: Randell Dawson <5313213+RandellDawson@users.noreply.github.com>

* Update guide/english/certifications/coding-interview-prep/data-structures/invert-a-binary-tree/index.md

Co-Authored-By: Randell Dawson <5313213+RandellDawson@users.noreply.github.com>

* Update guide/english/certifications/coding-interview-prep/data-structures/invert-a-binary-tree/index.md

Co-Authored-By: Randell Dawson <5313213+RandellDawson@users.noreply.github.com>

* Update guide/english/certifications/coding-interview-prep/data-structures/invert-a-binary-tree/index.md

Co-Authored-By: Randell Dawson <5313213+RandellDawson@users.noreply.github.com>

* fix: added extra line to avoid linting errors
2019-06-20 13:28:37 -04:00

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title
title
Invert a Binary Tree

Invert a Binary Tree

Hint: 1

Create a invert(node = this.root) method in the BinarySearchTree constructor function.

try to solve the problem now

Hint: 2

Try to use recursion and think of a base case.

try to solve the problem now

Spoiler Alert!

Solution ahead!

Basic Code Solution:

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
  this.invert = function(node = this.root) {
    if (node) {
      const temp = node.left;
      node.left = node.right;
      node.right = temp;
      this.invert(node.left);
      this.invert(node.right);
    }
    return node;
  }
    // change code above this line
}

Run Code

Code Explanation:

  • Using recursion will allow you to traverse each node once and the only extra memory used is the auxiliary temp variable that enables you to swap. You keep swapping the left and right pointers of a node until you reach the leaves which will not do anything as the left and right of them are null references.

NOTES FOR CONTRIBUTIONS:

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