一、题目

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。

二、解法

 1 /**
 2 public class TreeNode {
 3     int val = 0;
 4     TreeNode left = null;
 5     TreeNode right = null;
 6
 7     public TreeNode(int val) {
 8         this.val = val;
 9
10     }
11
12 }
13 */
14 public class Solution {
15     public TreeNode Convert(TreeNode pRootOfTree) {
16         if(pRootOfTree == null)
17             return null;
18         //如果是叶子结点
19         if(pRootOfTree.left == null && pRootOfTree.right == null)
20             return pRootOfTree;
21         //1、先从左子树遍历
22         TreeNode left = Convert(pRootOfTree.left);
23         TreeNode p = left;//p指向当前链表的最后一个结点
24         while(p != null && p.right != null)
25             p = p.right;
26         //2、与当前的pRootOfTree进行连接
27         //p.right = pRootOfTree; pRootOfTree.left = p;
28         if(left != null){
29             p.right = pRootOfTree;
30             pRootOfTree.left = p;
31         }
32         //3、右子树转换为双向链表
33         TreeNode right = Convert(pRootOfTree.right);
34         if(right != null){
35             pRootOfTree.right = right;
36             right.left = pRootOfTree;
37         }
38         return left != null ? left : pRootOfTree;
39     }
40 }