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

难度：medium

题目：给定二叉树中序及后序遍历，构造二叉树，注意二叉树中的结点不重复。

思路；分治。

1. 从后序遍历数组中由后向前取结点r
2. 在中序遍历中找到r的位置，并由此结点分成两部分，继续执行1.

Runtime: 4 ms, faster than 68.02% of Java online submissions for Construct Binary Tree from Inorder and Postorder Traversal.

Memory Usage: 37.6 MB, less than 42.45% of Java online submissions for Construct Binary Tree from Inorder and Postorder Traversal.

/** * 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 (null == postorder || postorder.length < 1) {            return null;        }                return buildTree(inorder, 0, inorder.length - 1, postorder, postorder.length - 1);    }        public TreeNode buildTree(int[] inorder, int start, int end, int[] postorder, int idx) {        if (start > end || idx < 0) {            return null;        }                   TreeNode root = new TreeNode(postorder[idx]);        int rIdx = start;        for (; rIdx <= end; rIdx++) {            if (inorder[rIdx] == postorder[idx]) {                break;            }        }                root.left = buildTree(inorder, start, rIdx - 1, postorder, idx - (end - rIdx) - 1);        root.right = buildTree(inorder, rIdx + 1, end, postorder, idx - 1);                return root;    }}