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

tag: binary tree, DFS

/**
 * 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) {
        if (preorder == null || preorder.length == 0) return null;
        Map<Integer, Integer> hash = new HashMap<>();
        for (int i = 0; i < inorder.length; i++){
            hash.put(inorder[i], i);
        }
        return dfs(0, 0, inorder.length - 1, hash, preorder, inorder);
    }

    private TreeNode dfs(int preStart, int inStart, int inEnd, Map<Integer, Integer> hash, int[] preorder, int[] inorder){
        if (preStart >= preorder.length || inStart > inEnd){
            return null;
        }

        TreeNode root = new TreeNode(preorder[preStart]);
        int inIndex = hash.get(preorder[preStart]);
        root.left = dfs(preStart + 1, inStart, inIndex - 1, hash, preorder, inorder);
        root.right = dfs(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, hash, preorder, inorder);

        return root;
    }
}