Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

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

Solution:
1. Construct root from the last element in post-order.
2. Find root's index(hash map) in in-order, and then we can decide its left sub-tree and right sub-tree.

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        // inorder value ---> index
        HashMap map = new HashMap();
        
        for(int i = 0; i < inorder.length; i++)
            map.put(inorder[i], i);
            
        return helper(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1, map);
    }
    
    // build up the tree recursively
    public TreeNode helper(int[] inorder, int i_beg, int i_end, int[] postorder, int p_beg, int p_end, HashMap map){
        if(i_beg > i_end || p_beg > p_end)
            return null;
        
        int root_val = postorder[p_end];
        TreeNode root = new TreeNode(root_val);
        
        int i = map.get(root_val);
        
        root.left = helper(inorder, i_beg, i-1, postorder, p_beg, p_beg + i-1-i_beg, map);
        root.right = helper(inorder, i+1, i_end, postorder, p_beg + i-1-i_beg + 1, p_end-1, map);
        
        return root;
    }
}