## Find if all nodes in BST are greater than a item

I have been working on trying to implement the function all_bigger below but I am not sure if there are flaws in my logic. To my understanding, BST's are organized having the smallest values on the left side so I would only need to check the left side of the BST. Is there a better way of writing this or is my code incorrect?

class BSTNode: """A node is a BST """ def __init__(self: 'BSTNode', item, left, right): self.item, self.left, self.right = item, left, right def all_bigger(self, value): """ >>> bst = BSTNode(5, BSTNode(4), BSTNode(6)) >>> all_bigger(bst, 2) True """ while self.left: if self.left > value: self.value = self.left: else: return False return True

Your code is almost correct, with some minor bugs. Corrected code:

class BSTNode: """A node is a BST """ def __init__(self, item, left = None, right = None): self.item, self.left, self.right = item, left, right def all_bigger(self, value): """ >>> bst = BSTNode(5, BSTNode(4), BSTNode(6)) >>> all_bigger(bst, 2) True """ root = self while(root!=None): if root.item > value: root = root.left else: return False return True bst = BSTNode(5, BSTNode(4,BSTNode(1, None, None),None), BSTNode(6,None,None)) # Returns False print(bst.all_bigger(2))

**Add all greater values to every node in a given BST,** Given a Binary Search Tree (BST), modify it so that all greater values in the given BST are added If we do reverse Inorder traversal of BST, we get all nodes in decreasing order. Node *newNode( int item) A wrapper over modifyBSTUtil(). Traverse node by node (Inorder, preorder, etc.) For each node find all the nodes greater than that of the current node, sum the values. Store all these sums. Replace each node value with their corresponding sum by traversing in the same order as in Step 1.

IIUC your question is to see if all the nodes in the BST are bigger than a certain value.

A simple way to do is to find the node with the minimum value in the BST and compare it with the other value. The smallest node is going to be the left-most node.

A typical BST node looks like this

# A binary tree node class Node: # Constructor to create a new node def __init__(self, key): self.data = key self.left = None self.right = None

And yes, you're right. The tree does not need to be searched fully, ie, the right subtrees can be skipped. This is how you find the node with the minimum value.

def minValue(node): current = node # loop down to find the lefmost leaf while(current.left is not None): current = current.left return current.data

This can be slightly tweaked to solve your problem

def all_bigger(node, val): current = node # loop down to find the lefmost leaf while(current.left is not None): current = current.left # Check if the current node value is smaller than val if current.data < val: return False return True

**Smallest number in BST which is greater than or equal to N ,** Otherwise, if node's value is greater than or equal to N and left child is NULL or less Else if node's value is less than N, then search for the element in the right If we were given a binary tree (not BST), then we need to traverse all nodes to find element. But, In case of BST, We are not required to traverse the all nodes of BST. BST has following properties. Left child of binary tree is less than its parent node; Right child of binary tree is greater than its parent node

You need to update the node after every comparison. Kindly check the below code:

class BSTNode: """A node is a BST """ def __init__(self: 'BSTNode', item, left, right): self.item, self.left, self.right = item, left, right def all_bigger(self, value): """ >>> bst = BSTNode(5, BSTNode(4), BSTNode(6)) >>> all_bigger(bst, 2) True """ while self.item: if self.left > value: self.item = self.left: else: return False return True

**How to check if a binary tree is a binary search tree,** If a node is a right child, then its key and the keys of the nodes in its left subtree are greater than its parent's key. Code. In the given tree, x = 7 Number of nodes greater than x are 4. Approach : The idea is maintain a count variable initialize to 0. Traverse the tree and compare root data with x. If root data is greater than x, increment the count variable and recursively call for all its children. Below is the implementation of idea.

**CS241: Data Structures & Algorithms II,** More terms are given on pages 445 - 447. Heap: a tree where parent node has bigger (smaller) value than children; And many others: binary search tree (BST), If a node c is the child of another node p, then we say that "p is c's parent". Each node of a binary tree can be stored as an object of a binary tree node class. Smallest number in BST which is greater than or equal to N. Given a Binary Search Tree and a number N, the task is to find the smallest number in the binary search tree that is greater than or equal to N. Print the value of the element if it exists otherwise print -1.

**Check if a binary tree is BST or not - Interview Problem,** All nodes in the left subtree of a node have values less than the node's if there is an element in the entire left subtree which is greater than the Add all greater values to every node in a given BST. Given a Binary Search Tree (BST), modify it so that all greater values in the given BST are added to every node. For example, consider the following BST. A simple method for solving this is to find sum of all greater values for every node.

**Binary Trees,** A binary tree is made of nodes, where each node contains a "left" pointer, to the node (<=), and all the elements in its right subtree are greater than the node (>). If the tree is a binary search tree, there is often some sort of less-than test on the In Java, we will have a BinaryTree object that contains a single root pointer. Largest number in BST which is less than or equal to N We have a binary search tree and a number N. Our task is to find the greatest number in the binary search tree that is less than or equal to N. Print the value of the element if it exists otherwise print -1.

**Binary search trees explained · YourBasic,** A binary search tree stores items in sorted order and offers efficient lookup, If p is a node and q is the root of p 's subtree, we say that p is the parent of q and that Every node in the right subtree of n contains a value which is larger than the A binary tree stores items that have a natural order in such a way that at each node X, all items stored in the left subtree of X are less than the item stored at X, and all items stored in the right subtree are greater than the item stored at X. Such a binary tree is called