[LC] 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.

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        Map<Integer, Integer> mymap = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            mymap.put(inorder[i], i);
        }
        return helper(0, inorder.length - 1, 0, preorder.length - 1, preorder, mymap);
    }
    
    private TreeNode helper(int inLeft, int inRight, int preLeft, int preRight, int[] preorder, Map<Integer, Integer> mymap) {
        if (inLeft > inRight) {
            return null;
        }
        TreeNode cur = new TreeNode(preorder[preLeft]);
        int leftSize = mymap.get(preorder[preLeft]) - inLeft; 
        // preRight for left just add up leftSize
        cur.left = helper(inLeft, inLeft + leftSize - 1, preLeft + 1, preLeft + leftSize, preorder, mymap);
        cur.right = helper(inLeft + leftSize + 1, inRight, preLeft + leftSize + 1, preRight, preorder, mymap);
        return cur;
    }
}