packageleetcode/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */// 解法一 非递归funcflatten(root*TreeNode){list,cur:=[]int{},&TreeNode{}preorder(roo
LeetCode—108. Convert Sorted Array to Binary Search Tree 题目 https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/description/ 将一个递增数组转化为平衡二叉搜索树。平衡二叉搜索树首先是一个二叉搜索树,其次要求每一个节点的左右... ...
* Source : https://oj.leetcode.com/problems/flatten-binary-tree-to-linked-list/ * * * 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 * \ *...
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
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,这里我们也用两种方法来求解。首先来看递归版本的...
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。
Given a binary tree, flatten it to a linked list in-place. For example, Given 1 / \ 2 5 / \ \ 3 4 6 1. 2. 3. 4. 5. The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 1. 2. 3. 4. 5. 6. 7. ...
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 ...
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 ...
也就是6作为5的右孩子,5作为4的右孩子,以此类推。 class Solution { private TreeNode prev = null; public void flatten(TreeNode root) { if (root == null) return; flatten(root.right); flatten(root.left); root.right = prev; root.left = null; prev = root; } }...