牛客网刷题之二叉搜索树与双向链表

来源：互联网发布：mac版战网客户端编辑：程序博客网时间：2024/06/18 04:29

题目描述：

这里写图片描述

解题思路：

　　二叉搜索树又名排序树，特点的是如果不为空，那么左子树必定小于根节点，而右子树必定大于根节点。题目要求转换为排序的双向链表，其实就相当于中序遍历该二叉搜索树但要加入双向指针：可以使后继结点的左指针指当前结点，同时当前结点的右指针指向后继，最后使后继结点等于当前结点。

题解：

//递归法TreeNode later = null;    public TreeNode Convert(TreeNode pRootOfTree) {        if (pRootOfTree == null) {            return null;        }        Convert(pRootOfTree.right);        if (later == null) {            later = pRootOfTree;        } else {            later.left = pRootOfTree;            pRootOfTree.right = later;            later = pRootOfTree;        }        Convert(pRootOfTree.left);        return later;    }

ac结果：

这里写图片描述
另附非递归法：

 public TreeNode Convert(TreeNode pRootOfTree) {        if (pRootOfTree == null) {            return null;        }        Stack<TreeNode> s = new Stack<>();        TreeNode later = null;        while(pRootOfTree != null || ! s.isEmpty()){            if(pRootOfTree != null){                s.push(pRootOfTree);                pRootOfTree = pRootOfTree.right;            }else{                pRootOfTree = s.pop();                if(later == null){                    later = pRootOfTree;                }else{                    later.left  = pRootOfTree;                    pRootOfTree.right = later;                    later = pRootOfTree;                }                pRootOfTree = pRootOfTree.left;            }        }        return later;    }

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