106. 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.

思路：
这题的思路与 105 Construct Binary Tree from Preorder and Inorder Traversal 基本相同。
不同点在于：
后序遍历中根的顺序是从后往前的。因此遍历后序数组时，应当从后往前搜索。
后序遍历时先经过右子树的根再经过左子树的根。因此，递归时先递归遍历右子树，再递归遍历左子树。最后把得到的两个子树接到当前的根结点上。

/**
 * 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 || postorder == null || postorder.length == 0 || inorder.length != postorder.length)
            return null;
        HashMap<Integer, Integer> map = new HashMap();
        for(int i = 0; i < inorder.length; i ++)
            map.put(inorder[i], i);
        return helper(inorder, 0, inorder.length - 1, postorder, postorder.length - 1, map);
    }

    TreeNode helper(int[] inorder, int startIn, int endIn, int[] postorder, int endPost, HashMap<Integer, Integer> map) {
        if(startIn > endIn)
            return null;
        TreeNode root = new TreeNode(postorder[endPost]);
        int index = map.get(root.val);
        root.right = helper(inorder, index + 1, endIn, postorder, endPost - 1, map);
        root.left = helper(inorder, startIn, index - 1, postorder, endPost - (endIn - index + 1), map);
        return