publicvoidflatten(TreeNode root){if(root ==null)return; flatten(root.left); flatten(root.right);if(root.left !=null){//左节点不为空,执行重连接操作TreeNoderight=root.right;//保存右节点root.right = root.left;//更新右节点为左节点root.left =null;//左节点置空TreeNodelast=root;//遍历新右...
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 ...
public void flatten(TreeNode root) { if (root == null) { return; } TreeNode[] list = new TreeNode[1]; helper(list, root); } public void helper(TreeNode[] list, TreeNode root) { if (root == null) { return; } TreeNode right = root.right; if (list[0] != null) { list[...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */classSolution{public:voidflatten(TreeNode*root){// Start typing your C/C++ solution below// DO NOT write int main() functionif(root==NULL)return;if(root->left)flatten(root->left);if(root->right)flatten(root->righ...
Given a binary tree, flatten it to a linked list in-place. For example, Given AI检测代码解析 1 / \ 2 5 / \ \ 3 4 6 1. 2. 3. 4. 5. The flattened tree should look like: AI检测代码解析 1 \ 2 \ 3 \ 4 \ 5 \ 6
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...
The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 难度:medium 题目:给定二叉树,原地扁平化。 思路:后序遍历 Runtime: 6 ms, faster than 100.00% of Java online submissions for Flatten Binary Tree to Linked List. Memory Usage: 40 MB, less than 100.00% of Java online submiss...
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...
//#109Description: Convert Sorted List to Binary Search Tree | LeetCode OJ 解法1:和之前一样,只不过这次改链表了。你可以凑活,也可以先转成数组。 // Solution 1: Same as before, only in that it's a list this time. You can either make do with it or convert it to array first. ...
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...