Convert Binary Search Tree to Doubly Linked List(将二叉查找树转换成双链表)

问题

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

return 1<->2<->3<->4<->5

分析

使用递归，注意是遍历的次序。自身的节点是new出来的，只有一个所以比较简单。左边和右边的都可能是一个链表，不是只有一个节点。所以在node指向右边的时候是没有问题的，在指向左边的时候需要把左边的链表遍历到尾节点。

代码

 * 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) {  
        if(root==null){
            return null;
        }
        DoublyListNode left=bstToDoublyList(root.left);
        DoublyListNode node=new DoublyListNode(root.val);
        DoublyListNode right=bstToDoublyList(root.right);
        node.next=right;
        if(right!=null){
            right.prev=node;
        }
        if(left!=null){
            DoublyListNode temp=left;
            while(true){
                if(temp.next==null){
                    temp.next=node;
                    node.prev=temp;
                    return left;
                }
                temp=temp.next;
            }
        }
        return node;
    }
}