题目描述：

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

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

解题思路:先序遍历的第一个元素为根元素，在中序遍历序列中找到根元素的位置。中序遍历根元素左边为左子树中序遍历序列，右边为右子树中序遍历序列。根据左右子树中序遍历序列的长度可以将先序遍历序列分割成左右两部分，左边为左子树的先序遍历序列，右边为右子树的先序遍历序列。此时左右子树的先序序列和中序序列都已知，对左右子树递归地调用上述方法，即可得结果。

代码：

TreeNode * build_tree(vector<int>::iterator prel,vector<int>::iterator prer,vector<int>::iterator inl,vector<int>::iterator inr)
{
    if(prel >= prer)
        return NULL;
    TreeNode * root;
    if(prer - prel == 1)
    {
        root = new TreeNode(0);
        root->val = *prel;
        return root;
    }
    root = new TreeNode(0);
    root->val = *prel;
    vector<int>::iterator root_pos = find(inl,inr,root->val);
    int left_length = root_pos - inl;
    root->left = build_tree(prel+1,prel+1+left_length,inl,inl+left_length);
    root->right = build_tree(prel+1+left_length,prer,inl+left_length+1,inr);
    return root;
}

TreeNode * Solution::buildTree(vector<int> &preorder, vector<int> &inorder)
{
    return build_tree(preorder.begin(),preorder.end(),inorder.begin(),inorder.end());
}

LeetCode:Construct Binary Tree from Preorder and Inorder Traversal