給定一棵樹的前序遍歷與中序遍歷，依據此構造二叉樹。
註意:
你可以假設樹中沒有重復的元素。
例如，給出
前序遍歷 = [3,9,20,15,7]
中序遍歷 = [9,3,15,20,7]
返回如下的二叉樹：
3
/ \
9 20
/ \
15 7
詳見：https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/

/**
 * 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) {
        int is=inorder.size();
        if(is==0||inorder.empty())
        {
            return nullptr;
        }
        int val=preorder[0];
        TreeNode* root=new TreeNode(val);
        vector<int> pre_left,pre_right,in_left,in_right;
        int p=0;
        for(;p<is;++p)
        {
            if(inorder[p]==val)
            {
                break;
            }
        }
        for(int i=0;i<is;++i)
        {
            if(i<p)
            {
                pre_left.push_back(preorder[i+1]);
                in_left.push_back(inorder[i]);
            }
            else if(i>p)
            {
                pre_right.push_back(preorder[i]);
                in_right.push_back(inorder[i]);
            }
        }
        root->left=buildTree(pre_left,in_left);
        root->right=buildTree(pre_right,in_right);
        return root;
    }
};

105 Construct Binary Tree from Preorder and Inorder Traversal 從前序與中序遍歷序列構造二叉樹