//Recursive solution
//Time: O(n), 3ms
//Space: O(h)
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        //Hash all the value in a HashTable so that we can look up the index in the inorder of a value in O(1) time
        Map<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < inorder.length; i++) {
            map.put(inorder[i], i);
        }
        return buildTreeR(inorder, postorder, 0, inorder.length, 0, postorder.length, map);
    }

    private TreeNode buildTreeR(int[] inorder, int[] postorder, int inBegin, int inEnd, int postBegin, int postEnd, Map<Integer, Integer> map) {
        if(inBegin == inEnd && postBegin == postEnd) return null;
        //The Last node of post-order is the root of the tree
        TreeNode root = new TreeNode(postorder[postEnd - 1]);
        int rootIndex = map.get(postorder[postEnd - 1]);
        //The left subtree and right subtree are divided by the root
        root.left = buildTreeR(inorder, postorder, inBegin, rootIndex, postBegin, postBegin + (rootIndex - inBegin), map);
        root.right = buildTreeR(inorder, postorder, rootIndex + 1, inEnd, postBegin + (rootIndex - inBegin), postEnd - 1, map);
        return root;
    }
}
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */