Convert a binary search tree to doubly linked list with in-order traversal.
Example
Given a binary search tree:
    4
   /  \
  2   5
  / \
1   3
return 1<->2<->3<->4<->5

我的思路还是追求逻辑最简单，思路最Straightforward但最不容易出错的“傻壮”方法。先求出inorder traversal的ArrayList, 然后将ArrayList转换成Doubly Linked List.

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Definition for Doubly-ListNode.
 * public class DoublyListNode {
 *     int val;
 *     DoublyListNode next, prev;
 *     DoublyListNode(int val) {
 *         this.val = val;
 *         this.next = this.prev = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param root: The root of tree
     * @return: the head of doubly list node
     */
    public DoublyListNode bstToDoublyList(TreeNode root) {  
        // Write your code here
        // brute force: first do inOrder traversal, then convert the 
        // list to a doubly linked list
        List<Integer> inOrder = new ArrayList<>();
        if (root == null){
            return null;    
        }
        TreeNode curt = root;
        Stack<TreeNode> stack = new Stack<>();
        while (curt != null){
            stack.push(curt);
            curt = curt.left;
        }
        while (!stack.isEmpty()){
            TreeNode node = stack.pop();
            inOrder.add(node.val);
            curt = node.right;
            while (curt != null){
                stack.push(curt);
                curt = curt.left;
            }
        }
        DoublyListNode res = new DoublyListNode(0);
        res.prev = null;
        DoublyListNode dummy = res;
        if (inOrder.size() == 1){
            return new DoublyListNode(inOrder.get(0));
        }
        for (int i = 0; i < inOrder.size() - 1; i++){
            DoublyListNode curtNode = new DoublyListNode(inOrder.get(i));
            DoublyListNode nextNode = new DoublyListNode(inOrder.get(i + 1));
            curtNode.next = nextNode;
            nextNode.prev = curtNode;
            res.next = curtNode;
            curtNode.prev = res;
            res = res.next;
        }
        return dummy.next;
    }
}