71：Construct Binary Tree from Preorder and Inorder Traversal

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

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>& preorder, vector<int>& inorder) {                return buildTree(begin(preorder), end(preorder), begin(inorder), end(inorder));        }        template <class InputIterator>        TreeNode* buildTree(InputIterator pre_first, InputIterator pre_last, InputIterator in_first, InputIterator in_last) {                if (pre_first == pre_last) return nullptr;                if (in_first == in_last) return nullptr;                auto root = new TreeNode(*pre_first);                auto inRootPos = find(in_first, in_last, *pre_first)                auto leftSize = distance(in_first, inRootPos);                auto pre_left_last = next(pre_first, leftSize + 1);                root -> left = buildTree(next(pre_first), pre_left_last, in_first, inRootPos);                root -> right = buildTree(pre_left_last, pre_last, next(inRootPos), in_last);                return root;        }};