72：Construct Binary Tree from Inorder and Postorder Traversal

题目：Given inorder and postorder traversal of a tree, construct the binary tree.
Note: You may assume that duplicates do not exist in the tree.

该题目解法的思想和上一题用前序和中序数据构造二叉树那道题目的解法一样

下面解法代码的思想及编写参考了网址https://github.com/soulmachine/leetcode#leetcode题解题目

代码如下：

/ 递归，时间复杂度 O(n)，空间复杂度 O(log n)class Solution {public:        TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {                return buildTree(begin(inorder), end(inorder), begin(postorder), end(postorder));        }        template <class InputIterator>        TreeNode* buildTree(InputIterator in_first, InputIterator in_last, InputIterator post_first, InputIterator post_last) {                if (in_first == in_last) return nullptr;                if (post_last == post_last) return nullptr;                const auto val = *prev(post_last);                auto root = new TreeNode(val);                auto inRootPos = find(in_first, in_last, val);                auto leftSize = distance(in_first, inRootPos);                auto post_left_last = next(post_first, left_size);                root -> left = buildTree(in_first, inRootPos, post_first, post_left_last);                root -> right = buildTree(next(in_root_pos), in_last, post_left_last, prev(post_last));                return root;        }};