二叉搜索树与双向链表

思路

题解

class Solution {
    Node pre, head;
    public Node treeToDoublyList(Node root) {
        if(root == null) return null;
        dfs(root);
        head.left = pre;
        pre.right = head;
        return head;
    }
    void dfs(Node cur) {
        if(cur == null) return;
        dfs(cur.left);
        if(pre != null) pre.right = cur;
        else head = cur;
        cur.left = pre;
        pre = cur;
        dfs(cur.right);
    }
}

二叉搜索树与双向链表

原文：https://www.cnblogs.com/treasury/p/12868437.html