``` 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 ...
要求:Given a binary tree, flatten it to a linked list in-place.将二叉树转化为平坦序列的树。比如: 结题思路: 该题有个提示,转化后的树的序列正好是二叉树前序遍历所得到的序列,所以,该题第一个思路就是利用前序遍历的方式来做。 第二个思路:我们可以利用递归的思路,先对根节点进行处理,将root的左子...
*/classSolution{public:voidflatten(TreeNode*root){// Start typing your C/C++ solution below// DO NOT write int main() functionbool flag=true;stack<TreeNode*>t;TreeNode*pre=NULL;if(root){t.push(root);}while(t.size()){TreeNode*cur=t.top();if(flag){if(cur->left&&cur->left!=pre...
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...
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 递归法 思路 相当于一次先序遍历, 将先访问的点存储在数组里, 下一个访问的节点为之前访问的点的右子树 ...
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...
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...