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Construct Binary Tree

Certification: Intermediate Level Accuracy: 0% Submissions: 0 Points: 10

Write a JavaScript program to construct a binary tree from its preorder and inorder traversal arrays. You may assume that duplicates do not exist in the tree. The preorder traversal visits nodes in the order: root, left subtree, right subtree. The inorder traversal visits nodes in the order: left subtree, root, right subtree.

Example 1
    
      
  • Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
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  • Output: [3,9,20,null,null,15,7]
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  • Explanation: 
    • The first element in preorder (3) is the root of the tree. 
    • Find 3 in inorder array at index 1.
    • Elements to the left of 3 in inorder ([9]) form the left subtree. 
    • Elements to the right of 3 in inorder ([15,20,7]) form the right subtree. 
    • Recursively construct left and right subtrees using the same logic.
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Example 2
    
      
  • Input: preorder = [-1], inorder = [-1]
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  • Output: [-1]
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  • Explanation: 
    • The preorder array has only one element (-1). 
    • This single element is the root of the tree. 
    • There are no left or right subtrees. 
    • The resulting tree has only one node with value -1.
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Constraints
    
      
  • 1 ≤ preorder.length ≤ 3000
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  • inorder.length == preorder.length
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  • -3000 ≤ preorder[i], inorder[i] ≤ 3000
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  • preorder and inorder consist of unique values
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  • Each value of inorder also appears in preorder
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  • Time Complexity: O(n)
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  • Space Complexity: O(n)
    •
Hash MapBinary TreeeBayApple
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Hint
Solution

Solution Hints

    
      
  • The first element in preorder is always the root of the current subtree
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  • Find the root's position in the inorder array to determine left and right subtrees
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  • Use a hash map to store inorder indices for O(1) lookup time
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  • Recursively construct left subtree using elements before root in inorder
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  • Recursively construct right subtree using elements after root in inorder
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  • Keep track of preorder index to get the next root for each recursive call
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Steps to solve by this approach:

 Step 1: Create a hash map to store the indices of elements in the inorder array for O(1) lookup.

 Step 2: Initialize a preorder index pointer to track the current root in the preorder traversal.

 Step 3: Create a recursive helper function that takes left and right boundaries of the current subtree.

 Step 4: Get the root value from the current preorder index and create a new TreeNode.

 Step 5: Find the root's position in the inorder array using the hash map.

 Step 6: Recursively build the left subtree using elements before the root in inorder.

 Step 7: Recursively build the right subtree using elements after the root in inorder.

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