[LeetCode]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.

这道题要求根据二叉树的前序遍历序列和中序遍历序列构建二叉树。
举个例子：
前序序列：A B D E F C H I
中序序列：D B E F A C I H
A是树的根节点，在中序序列中找到A 的位置，则中序序列中A左边的序列构成结点A的左子树，A右边的序列构成结点A的右子树。对左右子树分别重复上述方式。
下面贴上代码：

TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {        return build(preorder,0,preorder.size(), inorder,0,inorder.size());    }    TreeNode* build(vector<int>& preorder,int pleft,int pright, vector<int>& inorder,int ileft,int iright){        if (ileft==iright)            return NULL;        int i = ileft;        while (preorder[pleft] != inorder[i]){            i++;        }        TreeNode* tn = new TreeNode(inorder[i]);        tn->left =build(preorder, pleft + 1, pleft + i - ileft, inorder, ileft, i);        tn->right = build(preorder, pleft + i - ileft + 1, pright, inorder, i + 1, iright);        return tn;    }

一开始没注意两个vector 之前&，于是写出了Merry Limit Exceeded的代码QAQ，也贴上来吧：

class Solution {public:    vector<int> getPart(vector<int>& v, int begin, int len){        vector<int> ans;        if (begin < v.size()){            for (int i = 0; i < len; i++){                ans.push_back(v[i + begin]);            }        }        return ans;    }    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder){        return build(preorder, inorder);    }    TreeNode* build(vector<int> preorder, vector<int> inorder){        if (preorder.empty())            return NULL;        TreeNode* tn = new TreeNode(preorder[0]);        vector<int>::iterator it = find(inorder.begin(), inorder.end(), preorder[0]);        int mid = it - inorder.begin();        int left = it - inorder.begin();        int right = inorder.end() - 1 - it;        tn->left = buildTree(getPart(preorder, 1, left), getPart(inorder, 0, left));        tn->right = buildTree(getPart(preorder, left + 1, right), getPart(inorder, mid + 1, right));        return tn;    }};