An unbalanced tree 1 / 10 / 5. Given two arrays that represent Preorder traversals of a Full binary tree preOrder[] and its mirror tree preOrderMirror(), your task is to complete the function constructBinaryTree(), that constructs the full binary tree using these two Pre Platform to practice programming problems. 2(4) Published October 12, 2018 at 471 × 362 in 2(4). Example 1: Input: 5 / \ 1 1 / \ 2 2 Output: True Explanation: Tree is mirror image of itself i.e. A tree is height balanced if difference between heights of left and right subtrees is not more than one for all nodes of tree. Check whether it is Symmetric or not, i.e. Your task is to complete the function height() which takes root node of the tree as input parameter and returns an integer denoting the height of the tree. Platform to practice programming problems. whether the binary tree is a Mirror image of itself or not.. Binary Tree to BST Easy Accuracy: 50.0% Submissions: 16623 Points: 2 Given a Binary Tree, convert it to Binary Search Tree in such a way that keeps the original structure of Binary Tree intact. A height balanced tree 1 / \ 10 39 / 5. Given a binary tree, find its height.â€‹ Example 1: Input: 1 / \ 2 3 Output: 2 Example 2: Input: 2 \ 1 / 3 Output: 3 Your Task: You don't need to read input or print anything. Given a Binary Tree, convert it into its mirror. Example 1: Given a Binary Tree. Solve company interview questions and improve your coding intellect Given a binary tree, find if it is height balanced or not. Example 1: Input: 1 / \ 2 3 Output: 2 1 3 Explanation: The tree is 1 (mirror) 1 / \ => / \ 3 2 2 3 The inorder of mirror is 2 1 3 Example 2: Input: 10 / \ 20 30 / \ 40 60 Output: 30 10 60 20 40 Explanation: The tree is 10 10 / \ (mirror) / \ 20 30 => 30 20 / \ / \ 40 60 60 40 The inroder traversal of mirror is 30 10 60 20 40. Solve company interview questions and improve your coding intellect ← Previous Next → tree is symmetric Example 2: Input: 5 / \ 10 10 / \ \ 20 20 30 Output: False search close; Home; Courses; GBlog; Puzzles; What's New ?

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