Construct a Tree from Level-Order and Inorder Traversals

Given the levelOrder and inorder traversals of the same binary tree, reconstruct and return its root.
All node values are distinct. Both arrays contain the same values, and together they describe exactly one valid binary tree.

Function
buildTree(levelOrder: int[], inorder: int[]) → TreeNode

Examples
Example 1
levelOrder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
return = [3,9,20,null,null,15,7]
The root 3 appears first in level order; the inorder split places 9 left and the remaining nodes right.
Example 2
levelOrder = [1]
inorder = [1]
return = [1]
Both traversals describe a one-node tree.
Example 3
levelOrder = [1,2,3,4,5]
inorder = [4,2,5,1,3]
return = [1,2,3,4,5]
The traversals reconstruct the shown complete upper levels.

Constraints
1 <= levelOrder.length == inorder.length <= 1200.
-10^9 <= levelOrder[i], inorder[i] <= 10^9.
Each traversal contains distinct values, and both contain the same set of values.
The arrays are valid traversals of one binary tree.

The trick: inorder tells you which side of the root every value lives on, and level order tells you the order in which to pick roots. Build a hash map from value to inorder index. Recurse on an inorder range. At each call, you need the node from that range that appears earliest in level order. Simple approach: for each subtree, filter the level-order array to values in the inorder range, preserving order. The first element is the root. Split by its inorder index, then recurse left and right with the filtered lists. With n up to 1200, O(n^2) is fine. The pitfall is slicing level order by position like you would for preorder. That breaks, because left and right nodes are interleaved. Another pitfall is forgetting the empty range base case. If you freeze on the filtering step during the live OA, StealthCoder is the hedge that gets you unstuck.

The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.