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: recursion

这道题和Construct Binary Tree from Preorder and Inorder Traversal思路完全一样。
这里的区别是要从中序遍历和后序遍历中构造出树，算法还是一样，只是现在取根是从后面取（因为后序遍历根是遍历的最后一个元素）。将inorder[]放进hashmap里，然后根据根的值从inorder中取得对应的index。
时间复杂度和空间复杂度也还是O(n)。
( 根据preorder和postorder不能重新构造出树来， 不能唯一确定一棵树)

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null)
            return null;
        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);
    }
    
    public TreeNode helper(int[] inorder, int inL, int inR, int[] postorder, int posL, int posR, HashMap map){
        if(inL > inR || posL > posR)
            return null;
        
        TreeNode root = new TreeNode(postorder[posR]);
        int index = map.get(root.val);
        root.left = helper(inorder, inL, index-1, postorder, posL, posL+index-inL-1, map);
        root.right = helper(inorder, index+1, inR, postorder, posR-(inR-index), posR-1, map);
        return root;
    }
}