## Binary Tree Insert without ordering

Is there any way to insert a new node in a binary tree (not bst) without comparing key values? The following code only works for the very first three nodes.

node *insert (node *root, int *key) { if (root==NULL) { root=newNode(root, key); return root; } else if (root->left == NULL) root->left=insert(root->left,key); else if (root-> right == NULL) root->right=insert(root->right,key); return root; }

**Insertion in a Binary Tree in level order,** inorder(temp->right);. } /*function to insert element in binary tree */. void insert(Node* temp, int key). {. queue<Node*> q;. q.push(temp);. // Do level order traversal The left and right subtree each must also be a binary search tree. There must be no duplicate nodes. The above properties of Binary Search Tree provide an ordering among keys so that the operations like search, minimum and maximum can be done fast. If there is no ordering, then we may have to compare every key to search a given key.

You could just keep track of the height of each node, and always insert it into the side with fewer children:

node *insert (node *root, int *key) { if (root==NULL) { root=newNode(root, key); root->height = 0 } else if (root->left == NULL) { insert(root->left,key); } else if (root->right == NULL) { insert(root->right,key); } else if (root->left->height <= root->right->height) { insert(root->left,key); } else { insert(root->right,key); } root->height++ }

**Binary Search Tree without Natural Ordering,** There's orderable data (which may be unordered), but it can be sorted by a variety of sorting algorithms. All of these sorting algorithms depend on the ability to Given a binary tree and a key, insert the key into the binary tree at first position available in level order. Recommended: Please try your approach on {IDE} first, before moving on to the solution. The idea is to do iterative level order traversal of the given tree using queue .

Comparing values is actually irrelevant, the only think you need to do is set a pointer. Since you didn't specify any real requirements, one solution could be as follows:

Changing the signature a bit so now you have a pointer to an already allocated node:

void insertNode(node *&root, node *newNode) { if (root == NULL) { root = newNode; return; } if (root->left == NULL) { root-left = newNode; return; } helperInsert(root->left, newNode); return; }

This will set the head (assuming I got the signature right), and otherwise check the left child.

void helperInsert(node *it, node *newNode) { if (it->left == NULL) { it->left = newNode; return; } helperInsert(it->left, newNode); return; }

This is obviously a flawed approach (the tree will not be balanced at the slightest), almost treating the tree as a linked list, but to my best understanding of the question, this is an example of how it can be done.

**Binary search tree,** An ordered binary tree is sometimes called a binary search tree. that it is possible to do a Record insertion into a tree without resetting any links to subtrees! Insertion in binary search tree. The structure and placement of each node depends on the order it is inserted into binary search tree. However, every insertion should leave binary search tree in correct state. A new node is added to binary search tree based on value. If the node is very first node to added to BST, create the node and make it root.

In

else if (root->left == NULL) root->left=insert(root->left,key);

you know `root->left`

is NULL so why to do the recursive call ?

Of course same for the next *else*

The following code only works for the very first three nodes.

If both *left* and *right* are non NULL you do not insert, that time it was necessary to do the recursive call on one of the two branches, and you will consider the key (so insert ordered) to decide which one. Note that the 2 tests to NULL you did are not correct if you insert to have a sorted tree ...

**Ordered trees,** A null pointer represents a binary tree with no elements -- the empty tree. A "binary search tree" (BST) or "ordered binary tree" is a type of binary tree where the I'm working on code for insertion into a binary search tree. It works for the first node I insert, making it the root, but after that it doesn't seem to insert any nodes. I'm sure it's a problem with setting left/right references, but I can't quite figure it out.

The heap advice is most sound. You don't need to heapify anything, just follow the rules that an element at index `k`

has children at `2*k + 1`

and `2*k + 2`

.

Another approach, useful when there is no array, but the nodes are generated on the fly, is to fill the tree level-wise. Notice that at level `k`

there are `2 ** k`

slots, conveniently indexed by a k-bit integer. Interpret the index as a path down the tree (clear bit tells to follow left child, set bit tells to follow a right one), along the lines of:

void insert_node(struct node * root, struct node * node, unsigned mask, unsigned path) { while ((mask >> 1) != 1) { root = mask & path? root->right: root->left; } if (mask & path) { assert (root->right == NULL); root->right = node; } else { assert (root->left == NULL); root->left = node; } } void fill_level_k(struct node * root, unsigned k) { unsigned slots = 1 << k; for (unsigned slot = 0; slot < slots; slot++) { struct node * node = generate_new_node(); insert_node(root, node, slots, slot); } }

**Binary Trees,** In a totally ordered binary tree, an inorder traversal encounters the vertices in This is similar to the Get operation, without the need to insert if the item is not A Better Solution is to augment every tree node to store count together with regular fields like key, left and right pointers. Insertion of keys 12, 10, 20, 9, 11, 10, 12, 12 in an empty Binary Search Tree would create following. 12 (3) / \ 10 (2) 20 (1) / \ 9 (1) 11 (1) Count of a key is shown in bracket

**12. Binary Search Trees,** Due to the way nodes in a binary search tree are ordered, an in-order traversal (left node, then root node, then right node) will always produce a sequence of You’ll have a degenerate binary tree, but if that bothers you then you can try the following: Use a DFS or BFS with a bounded depth to search for a node with a null child. If a null child is found insert the new key at this null point. If no free spaces are found then insert at root.

**CS 225,** Already in tree – so return false; do not insert again delete from left subtree (unless there is no left subtree) Option 2: sorted array with binary searching. The main operations in binary tree are: search, insert and delete. We will see the worst case time complexity of these operations in binary trees. Binary Tree – In a binary tree, a node can have maximum two children. Consider the left skewed binary tree shown in Figure 1.

**[PDF] BST search efficiency Insert to a BST Insertion order affects the tree ,** An important special kind of binary tree is the binary search tree (BST). postorder, inorder, or level-order) visits the nodes of a BST in sorted order? class definitions assume that the BST will store only key values, no associated data. Insert them in pseudo-random order, like here: Browse other questions tagged algorithm sorting tree binary-tree binary-search-tree or ask your own question.