Convert Binary Search Tree to Doubly Linked List

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

中序遍历 穿针引线

/**
 * 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 List<DoublyListNode> helper(List<DoublyListNode> hT, TreeNode root) {
        if (root.left != null) {
            helper(hT, root.left);
        }
        DoublyListNode newNode  =  new DoublyListNode(root.val);
        newNode.prev = hT.get(1);
        hT.get(1).next = newNode;
        hT.remove(1);
        hT.add(newNode);
        if (root.right != null) {
            helper(hT, root.right);
        }
        return hT;
    }
    public DoublyListNode bstToDoublyList(TreeNode root) {
        // Write your code here
        if (root == null) {
            return null;
        }
        DoublyListNode head = new DoublyListNode(-1);
        DoublyListNode tail = head;
        List<DoublyListNode> hT = new ArrayList<DoublyListNode>();
        hT.add(head);
        hT.add(tail);
        helper(hT, root);
        hT.get(0).next.prev = null;
        return hT.get(0).next;
    }
}