根據一棵樹的前序遍歷與中序遍歷構造二叉樹。

註意:
你可以假設樹中沒有重復的元素。

例如，給出

前序遍歷 preorder = [3,9,20,15,7]
中序遍歷 inorder = [9,3,15,20,7]

返回如下的二叉樹：

    3
   /   9  20
    /     15   7

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* dfs(vector<int>& preorder, vector<int>& inorder, int pre, int inl, int inr){
13         if(inl == inr) return new TreeNode(preorder[pre]);
14         if(inl > inr) return NULL;
15         TreeNode* root = new TreeNode(preorder[pre]);
 
16         int i = inl;
17         while(preorder[pre] != inorder[i] && i <= inr) i++;
18         root->left = dfs(preorder, inorder, pre+1, inl, i-1);
19         root->right = dfs(preorder, inorder, pre+i-inl+1, i+1, inr);
20         return root;
21     }
22     TreeNode* buildTree(vector<int
 
>& preorder, vector<int>& inorder) {
23         int l = preorder.size()-1;
24         return dfs(preorder, inorder, 0, 0, l);
25     }
26 };

leetcode 105. 從前序與中序遍歷序列構造二叉樹