26、二叉树与双向链表

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。

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

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

    }

}
*/
public class Solution {
    
    public TreeNode Convert(TreeNode pRootOfTree) {
        if(pRootOfTree == null){
            return null;
        }
        //左子树转换
        TreeNode head1 = Convert(pRootOfTree.left);
        if(head1!=null){
            //找左子树最后一个节点
            TreeNode lst = findLastNode(head1);
            //链接左子树、根
            lst.right = pRootOfTree;
            pRootOfTree.left = lst;
        }
        
        //右子树转换
        TreeNode head2 = Convert(pRootOfTree.right);
        if(head2!=null){
            //链接右子树、根
            pRootOfTree.right = head2;
            head2.left = pRootOfTree;
        }
        
        if(head1!=null){
            return head1;
        }else{
            return pRootOfTree;
        }
    }
    
    public TreeNode findLastNode(TreeNode head){
        TreeNode lst = head;
        while(lst.right!=null){
            lst = lst.right;
        }
        return lst;
    }
}