105. Construct Binary Tree from Preorder and Inorder Traversal

/**
 * 105. Construct Binary Tree from Preorder and Inorder Traversal
 *
 * https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/
 *
 * Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

3
/ \
9  20
/  \
15   7
 * */

?

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class Solution {       class TreeNode(var `val`: Int = 0) {         var left: TreeNode? = null;         var right: TreeNode? = null;     }       fun buildTree(preorder: IntArray, inorder: IntArray): TreeNode? {         if (preorder.isEmpty() || inorder.isEmpty())             return null;         val map = HashMap<Int, Int>();         for ((index, element) in inorder.withIndex()) {             map.put(element, index);         }         return help(preorder, 0, preorder.size - 1, inorder, 0, inorder.size - 1, map);     }       fun help(         preorder: IntArray,         pLeft: Int,         pRight: Int,         inorder: IntArray,         iLeft: Int,         iRight: Int,         map: HashMap<Int, Int>     ): TreeNode? {         if (pLeft > pRight || iLeft > iRight)             return null;         val node = TreeNode(preorder[pLeft]);         val rootIndex:Int = map.get(preorder[pLeft])!!;//not-null assertion operator:!! convert Int? to Int         node.left = help(preorder, pLeft+1, pLeft+rootIndex-iLeft,inorder,iLeft,rootIndex-1,map);         node.right = help(preorder, pLeft+rootIndex-iLeft+1,pRight,inorder,rootIndex+1,iRight,map);         return node;     } }