Construct Binary Tree from Preorder and Inorder Traversal<leetcode>

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

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

题意：知道二叉树的前序遍历和中序遍历，重新构建出二叉树。

算法：（略），代码如下：

/**
 * Definition for binary tree
 * 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) {
        if(preorder.empty()||inorder.empty())  return NULL;
        return build(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
    }
    TreeNode  *build(vector<int> &preorder,int l,int r,vector<int> &inorder,int ll,int rr)
    {
        if(l>r||ll>rr)  return NULL;
        TreeNode* root=new TreeNode(0);
        root->val=preorder[l];
        if(l==r)  return root;
        int mid=0;
        for(int i=ll;i<=rr;i++)
        {
            if(inorder[i]==preorder[l])
            {
                mid=i;
                break;
            }
        }
        int midd=mid-ll+l;
        root->left=build(preorder,l+1,midd,inorder,ll,mid-1);
        root->right=build(preorder,midd+1,r,inorder,mid+1,rr);
        return root;
    }
};