``` import java.util.ArrayList; import java.util.Arrays; import java.util.List; /** * Source : https://oj.leetcode.com/problems/flatten-binary-tree-to
Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Hints: If you notice carefully in the flattened tree, each node's right child points to the next node of ...
Can you solve this real interview question? Flatten Binary Tree to Linked List - Given the root of a binary tree, flatten the tree into a "linked list": * The "linked list" should use the same TreeNode class where the right child pointer points to the
[LeetCode] Flatten Binary Tree to Linked List 将二叉树展开成链表
Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 思路: 递归处理每个结点,如果结点左孩子为空,则无需处理。否则先将结点右孩子挂到左孩子最右边的子孙结点,然后将左...
Given a binary tree, flatten it to a linked list in-place. For example, Given 代码语言:javascript 代码运行次数:0 运行 AI代码解释 1/\25/\ \346 The flattened tree should look like: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 ...
click to show hints. Hints: If you notice carefully in the flattened tree, each node's right child points to the next node of a pre-order trave 这道题要求把二叉树展开成链表,根据展开后形成的链表的顺序分析出是使用先序遍历,那么只要是数的遍历就有递归和非递归的两种方法来求解,这里我们也用两种...
Right = rightHead tail = rightTail } // 返回当前子树转换成的链表头结点和尾结点 return head, tail } 题目链接: Flatten Binary Tree to Linked List: leetcode.com/problems/f 二叉树展开为链表: leetcode.cn/problems/fl LeetCode 日更第 195 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list/ 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。解法一：递归使用递归的方式求解，递归过程如下：如果当前根节点为null时，直接返回；如果当前根节点的左右子节点都为空时，不需要调整...
114Flatten Binary Tree to Linked ListC 113Path Sum IIC++ 112Path SumC 111Minimum Depth of Binary TreeC 110Balanced Binary TreeC 109Convert Sorted List to Binary Search TreeC++ 108Convert Sorted Array to Binary Search TreeC 107Binary Tree Level Order Traversal IIC++ ...