Construct Binary Tree from Preorder and Inorder Traversal

Problem statement

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]Output: [-1]

Constraints:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder and inorder consist of unique values.
• Each value of inorder also appears in preorder.
• preorder is guaranteed to be the preorder traversal of the tree.
• inorder is guaranteed to be the inorder traversal of the tree.

My solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {number[]} preorder
 * @param {number[]} inorder
 * @return {TreeNode}
 */
var buildTree = function(preorder, inorder) {
    const inOrderLength = inorder.length - 1;
    let root = traverse(inorder, preorder, 0, inOrderLength, 0)
    
    return root;
};

// 0 0 4 => 1 => 0 0 | 2 4 | 0 + 1 - 0 + 1
// 

function traverse(inOrder, preOrder, inOrderStart, inOrderEnd, preIndex) {
    // console.log(inOrderStart, inOrderEnd, preIndex)
    if (inOrderStart > inOrderEnd || preIndex > preOrder.length - 1) {
        return null;
    }
    
    const node = new TreeNode(preOrder[preIndex])
    
    // console.log(node.val)
    
    if (inOrderStart === inOrderEnd) {
        return node
    }
    
    const idx = search(inOrder, inOrderStart, inOrderEnd, node.val)
    
    node.left = traverse(inOrder, preOrder, inOrderStart, idx - 1, preIndex + 1)
    node.right = traverse(inOrder, preOrder, idx + 1, inOrderEnd, preIndex + idx - inOrderStart + 1)
    
    return node
}

function search(inOrder, left, right, target) {
    let i;
    for (i = left; i <= right; i++) {
        if (inOrder[i] === target) {
            return i
        }
    }
    return i;
}