Convert Binary Search Tree to Doubly Linked List

Convert a binary search tree to doubly linked list with in-order traversal.

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Example

Given a binary search tree:

    4
   / 
  2   5
 / 
1   3

return 1<->2<->3<->4<->5

Runtime: 15ms

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 * Definition of Doubly-ListNode
 * class DoublyListNode {
 * public:
 *     int val;
 *     DoublyListNode *next, *prev;
 *     DoublyListNode(int val) {
 *         this->val = val;
           this->prev = this->next = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param root: The root of tree
     * @return: the head of doubly list node
     */
    DoublyListNode* bstToDoublyList(TreeNode* root) {
        // Write your code here
        
        // do inorder traversal first
        vector<int> inorderStore;
        inorder(root, inorderStore);
        
        // use the values in vector to construct a double linked list
        DoublyListNode* pre= new DoublyListNode(0);
        return constructDoubleList(inorderStore, pre);
    }
private:
    void inorder(TreeNode* root, vector<int>& inorderStore) {
        if(!root) return;
        inorder(root->left, inorderStore);
        inorderStore.push_back(root->val);
        inorder(root->right, inorderStore);
    }
    
    DoublyListNode* constructDoubleList(vector<int>& inorderStore, DoublyListNode* pre) {
        if(inorderStore.empty()) return NULL;
        
        DoublyListNode* move = pre;
        for (int i = 0; i < inorderStore.size(); i++) {
            DoublyListNode* temp = new DoublyListNode(inorderStore[i]);
            move->next = temp;
            move = move->next;
        }
        
        DoublyListNode* head = pre->next;
        head->prev = NULL;
        return head;
    }
};