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 ...
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...
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[...
LeetCode 114题的解题思路是什么? 二叉树扁平化成链表的算法有哪些? 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。
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...
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 ...
[LeetCode]Flatten Binary Tree to Linked List Question 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:...
https://leetcode.com/problems/flatten-binary-tree-to-linked-list/ 题目: 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 ...
publicvoidPrintBinaryTreeBacRecur(TreeNode<T>root){if(root==null)return;PrintBinaryTreeBacRecur(root.right);PrintBinaryTreeBacRecur(root.left);System.out.print(root.data);} 这里的话,我们不再是打印根节点,而是利用一个全局变量pre,更新当前根节点的右指针为pre,左指针为null。
createTree(root->left); createTree(root->right); } } };intmain() { Solution s; TreeNode*root; s.createTree(root); s.flatten(root);while(root) { cout<<root->val<<""; root=root->right; } } 运行结果: 提交到leetcode时,一直报错,开始不太了解,后来才发现是因为左指针一直没有设为NUL...