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 click to show hints. Hints: If you notice carefully in the flattened tree, each node's right…

题目 给定一个二叉树,原地将它展开为链表. 例如,给定二叉树 1 / \ 2 5 / \ \ 3 4 6 将其展开为: 1 \ 2 \ 3 \ 4 \ 5 \ 6 解析 通过递归实现:可以用先序遍历,然后串成链表 主要思想就是:先递归对右子树进行链表化并记录,然后将root->right指向 左子树进行链表化后的头结点,然后一直向右遍历子树,连接上之前的右子树 /** * Definition for a binary tree node. * struct TreeNode { * int v…

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 click to show hints. Hints: If you notice carefully in the flattened tree, each node's right…

Flatten a binary tree to a fake "linked list" in pre-order traversal. Here we use the right pointer in TreeNode as the next pointer in ListNode. Notice Don't forget to mark the left child of each node to null. Or you will get Time Limit Exceeded…

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 \ 6Hints: If you notice carefully in the flattened tree, each node's right child points to the n…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 给一个二叉树,把它展平为链表 in-place 根据展平后的链表的顺序可以看出是先序遍历的结果,所以用inorder traversal. 解…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 题目 将左子树所形成的链表插入到root和root->right之间 思路 1 / 2(root) 假设当前root为2 / \ 3(p) 4…

/* 先序遍历构建链表,重新构建树 */ LinkedList<Integer> list = new LinkedList<>(); public void flatten(TreeNode root) { preOrder(root); TreeNode res = root; list.poll(); while (!list.isEmpty()) { res.right = new TreeNode(list.poll()); res.left = null; res =…

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 这道题就是将数变为只有右节点的树. 递归,不是很难. 递归的时候求出左节点的最右孩子即可. /** * Definition for a binary tree node. * pub…

Given a binary tree, flatten it to a linked list in-place. For example, given the following tree: 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 这个题思路就是DFS, 先左后右, 记住如果是用stack如果需要先得到左, 那么要先append右, 另外需要注意的就是用recursive…