[Leetcode]105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {        return buildTree(begin(preorder), end(preorder), begin(inorder), end(inorder));    }        TreeNode* buildTree(vector<int>::iterator pre_first, vector<int>::iterator pre_last, vector<int>::iterator in_first, vector<int>::iterator 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 leftLen = distance(in_first, inrootPos);        root->left = buildTree(next(pre_first), next(pre_first, leftLen + 1), in_first, next(in_first, leftLen));        root->right = buildTree(next(pre_first, leftLen + 1), pre_last, next(inrootPos), in_last);        return root;    }};