一、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。
二、思路
中序遍历树，并时时更新最后一个节点。
根据这三个节点的关系，判断出全局关系，从而写出代码。
三、代码实现
/**
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
*/
public class Solution {
TreeNode lastNodeInList = null;
public TreeNode ConvertNode(TreeNode node,TreeNode lastNodeInList){
if(node == null){
return null;
}
TreeNode current = node;
if(current.left != null){
lastNodeInList = ConvertNode(current.left,lastNodeInList);
}
current.left = lastNodeInList;
if(lastNodeInList != null){
lastNodeInList.right = current;
}
lastNodeInList = current;
// System.out.println(lastNodeInList.val);
if(current.right != null){
lastNodeInList = ConvertNode(current.right,lastNodeInList);//对于递归，必须要接一下返回的参数
}
return lastNodeInList;
}
public TreeNode Convert(TreeNode pRootOfTree) {
lastNodeInList = ConvertNode (pRootOfTree,lastNodeInList);
/*if(lastNodeInList == null) {
System.out.println("nihao");
}*/
TreeNode pHeadOfList = lastNodeInList;
while(pHeadOfList!=null && pHeadOfList.left!=null){
// System.out.println("nihoa");
pHeadOfList = pHeadOfList.left;
}
return pHeadOfList;
}
}