二叉搜索树与双向链表

题目描述：
输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。

import java.util.Stack;public class ConvertBSTreeToOrderBNode {    //非递归方法    public TreeNode Convert(TreeNode pRootOfTree) {        if(pRootOfTree == null){            return null;        }        Stack<TreeNode> stack = new Stack<TreeNode>();        TreeNode p = pRootOfTree;        Boolean isFirst = true;        TreeNode pHead = null;//排序的双向链表的头节点        TreeNode pre = pHead;// 记录前一个节点        while(p != null || !stack.isEmpty()){            while(p != null){                stack.push(p);                p = p.left;            }            p = stack.pop();            if(isFirst){                pHead = p;                pre = p;                isFirst = false;            }            else{                pre.right = p;                p.left = pre;                pre = p;            }            p = p.right;            }        return pHead;    }    //递归方法    public TreeNode Convert1(TreeNode pRootOfTree){        if(pRootOfTree == null){            return null;        }        if(pRootOfTree.left == null && pRootOfTree.right == null){            return pRootOfTree;        }        // 1.将左子树构造成双链表，并返回链表头节点        TreeNode leftHead = Convert(pRootOfTree.left);        TreeNode p = leftHead;        // 2.定位至左子树双链表最后一个节点        while(p != null && p.right != null){            p = p.right;        }        // 3.如果左子树链表不为空的话，将当前root追加到左子树链表        if(leftHead != null){            pRootOfTree.left = p;            p.right = pRootOfTree;        }        // 4.将右子树构造成双链表，并返回链表头节点        TreeNode rightHead = Convert(pRootOfTree.right);         // 5.如果右子树链表不为空的话，将该链表追加到root节点之后        if(rightHead != null){            rightHead.left = pRootOfTree;            pRootOfTree.right = rightHead;        }        return leftHead != null ? leftHead : pRootOfTree;    }}