二叉搜索树与双向链表

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。

/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/

思路

中序遍历递归
先左子树后右子树
设置一个移动指针,设置与相邻节点的左右指向,移动指针

链接：https://www.nowcoder.com/questionTerminal/947f6eb80d944a84850b0538bf0ec3a5
来源：牛客网

//直接用中序遍历
public class Solution {
    TreeNode head = null;
    TreeNode realHead = null;
    public TreeNode Convert(TreeNode pRootOfTree) {
        ConvertSub(pRootOfTree);
        return realHead;
    }

    private void ConvertSub(TreeNode pRootOfTree) {
        if(pRootOfTree==null) return;
        ConvertSub(pRootOfTree.left);
        if (head == null) {
            head = pRootOfTree;
            realHead = pRootOfTree;
        } else {
            head.right = pRootOfTree; //改变指向
            pRootOfTree.left = head;
            head = pRootOfTree;      //移动指针
        }
        ConvertSub(pRootOfTree.right);
    }
}