In this article, we are going to seehow we can convert a binary search tree into a greater sum tree?Before discussing the solution, let’s understand what does mean by a greater sum tree. A greater sum tree mean
*right;13node() : data(0), left(NULL), right(NULL) { }14node(intd) : data(d), left(NULL), right(NULL) { }15};1617voidprints(node *root) {18if(!root)return;19prints
Learn how to convert a binary tree in C++ so that every node stores the sum of all nodes in its right subtree with this comprehensive guide.
Given an integer arraynumswhere the elements are sorted inascending order, convertit to aheight-balancedbinary search tree. Example 1: Input:nums = [-10,-3,0,5,9]Output:[0,-3,9,-10,null,5]Explanation:[0,-10,5,null,-3,null,9] is also accepted: Example 2: Input:nums = [1,3]...
虽然说树类问题首推递归解法,但是中序遍历是可以用迭代来写的,可以参见博主之前的博客Binary Tree Inorder Traversal。迭代写法借用了栈,其实整体思路和递归解法没有太大的区别,递归的本质也是将断点存入栈中,以便之后可以返回,这里就不多讲解了,可以参见上面的讲解,参见代码如下: ...
题目描述(简单难度) 给一个升序数组,生成一个平衡二叉搜索树。平衡二叉树定义如下: 它是一棵空树或它的左右两个子树的高度差的绝对值不超过1,并且左右两个子树都是一棵平衡二叉树。 二叉搜索树定义如下: 若…
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1. Example: Given the sorted linked list: [-10,-3,0,5,9], One possible answer is: [0,-3,9,-10,null,5], which represent...
* Convert Sorted Array to Binary Search Tree */ public TreeNode sortedArrayToBST(int[] nums) { if (nums == null || nums.length == 0) { return null; } return test(nums, 0, nums.length - 1); } public TreeNode test(int[] nums, int left, int right) { ...
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.Example:1.一看之下感觉和108很像，后来发现是linklist，就又不会做了=。=于是在youtube上找到了公瑾讲解 2.这道题...
classSolution{public:TreeNode*sortedListToBST(ListNode*head){returncreate(head,NULL);}TreeNode*create(ListNode*head,ListNode*tail){if(head==tail)returnNULL;ListNode*slow=head,*fast=head;while(fast!=tail&&fast->next!=tail)slow=slow->next,fast=fast->next->next;TreeNode*root=newTreeNode(slow-...