LeetCode 105: Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // By preorder, we can obtain the root node.
        // By root node, we can divide inorder into two parts.
        // By the two parts, we can also divide preorder into two parts.
        // Recursively do this work.
        if (inorder.length==0 || inorder.length!=preorder.length)
            return null;
            
        int len = inorder.length;
        
        // Save <inorder number, inorder index>, so it will be fast to get the index position.
        Map<Integer, Integer> in = new HashMap<>();
        
        for (int i = 0; i < len; i++)
            in.put(inorder[i], i);
            
        int loIn = 0, hiIn = len-1;
        int loPre = 0, hiPre = len-1;
        
        return genTree(inorder, loIn, hiIn, preorder, loPre, hiPre, in);
    }
    
    private TreeNode genTree(int[] inorder, int loIn, int hiIn, int[] preorder, int loPre, int hiPre, Map<Integer, Integer> in)
    {
        if (loIn > hiIn)
            return null;
            
        TreeNode node = new TreeNode(preorder[loPre]);
        
        int id = in.get(preorder[loPre]);
        
        // New 'hiPre' = 'loPre' + ('hiIn' - 'loIn')    
        node.left = genTree(inorder, loIn, id-1, preorder, loPre+1, loPre+id-loIn, in);
        node.right = genTree(inorder, id+1, hiIn, preorder, loPre+id-loIn+1, hiPre, in);
        
        return node;
    }
}