算法描述：

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

解题思路：理解原理。

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return helper(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
    }
    
    TreeNode* helper(vector<int>& preorder, int ps, int pe, vector<int>& inorder, int is, int ie){
        if(ps > pe || is > ie) return nullptr;
        TreeNode* node = new TreeNode(preorder[ps]);
        int pos = 0;
        for(int i=is; i <= ie; i++){
            if(inorder[i]==preorder[ps]){
                pos = i;
                break;
            }
        }
        
        node->left = helper(preorder, ps+1, pe, inorder, is, pos-1);
        node->right = helper(preorder,ps+pos-is+1, pe, inorder, pos+1, ie);
        return node;
    }