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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

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[] inorder, int[] postorder) {
        if (inorder == null || inorder.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, postorder.length - 1, 0, inorder.length - 1, hash, inorder, postorder);
    }

    private TreeNode dfs(int postStart, int postEnd, int inStart, int inEnd, Map<Integer, Integer> hash, int[] inorder, int[] postorder){
        if (postStart > postEnd || inStart > inEnd) return null;

        TreeNode root = new TreeNode(postorder[postEnd]);
        int inIndex = hash.get(postorder[postEnd]);

        root.left = dfs(postStart, postStart + inIndex - inStart - 1, inStart, inIndex - 1, hash, inorder, postorder);
        root.right = dfs(postStart + inIndex - inStart, postEnd - 1, inIndex + 1, inEnd, hash, inorder, postorder);

        return root;
    }
}