一、題目

輸入一棵二叉搜索樹，將該二叉搜索樹轉換成一個排序的雙向鏈表。要求不能創建任何新的結點，只能調整樹中結點指針的指向。

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         //如果是葉子結點
19         if(pRootOfTree.left == null && pRootOfTree.right == null)
20             return pRootOfTree;
21         //1、先從左子樹遍歷
22         TreeNode left = Convert(pRootOfTree.left);
 
23         TreeNode p = left;//p指向當前鏈表的最後一個結點
24         while(p != null && p.right != null)
25             p = p.right;
26         //2、與當前的pRootOfTree進行連接
27         //p.right = pRootOfTree; pRootOfTree.left = p;
28         if(left != null){
29             p.right = pRootOfTree;
30             pRootOfTree.left = p;
 
31         }
32         //3、右子樹轉換為雙向鏈表
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 }

26、二叉搜索樹與雙向鏈表