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

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Yes

Example

Given a binary search tree:

    4
   / \
  2   5
 / \
1   3

return1<->2<->3<->4<->5

/**
 * 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
        if (root == nullptr)
            return nullptr;

        DoublyListNode* left = bstToDoublyList(root->left);
        DoublyListNode* right = bstToDoublyList(root->right);

        DoublyListNode* mid = new DoublyListNode(root->val);
        mid->next = right;
        if (right != nullptr)
            right->prev = mid;

        DoublyListNode head;
        head.next = left;
        DoublyListNode* cur = &head;
        while (cur->next != nullptr)
            cur = cur->next;
        cur->next = mid;
        mid->prev = cur;

        return head.next;
    }
};