Check whether a given Binary Tree is Complete or not | GeeksforGeeks
Given a Binary Tree, write a function to check whether the given Binary Tree is Complete Binary Tree or not.
A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible. See following examples.
Time Complexity: O(n) where n is the number of nodes in given Binary Tree
Auxiliary Space: O(n) for queue.
The approach is to do a level order traversal starting from root. In the traversal, once a node is found which is NOT a Full Node, all the following nodes must be leaf nodes.
Also, one more thing needs to be checked to handle the below case: If a node has empty left child, then the right child must be empty.
http://www.ideserve.co.in/learn/check-whether-binary-tree-is-complete-tree-or-not
Time Complexity is O(n)
Space Complexity is O(n)
http://www.geeksforgeeks.org/check-whether-binary-tree-complete-not-set-2-recursive-solution/
Read full article from Check whether a given Binary Tree is Complete or not | GeeksforGeeks
Given a Binary Tree, write a function to check whether the given Binary Tree is Complete Binary Tree or not.
A complete binary tree is a binary tree in which every level, except possibly the last, is completely filled, and all nodes are as far left as possible. See following examples.
Time Complexity: O(n) where n is the number of nodes in given Binary Tree
Auxiliary Space: O(n) for queue.
The approach is to do a level order traversal starting from root. In the traversal, once a node is found which is NOT a Full Node, all the following nodes must be leaf nodes.
Also, one more thing needs to be checked to handle the below case: If a node has empty left child, then the right child must be empty.
bool isCompleteBT(struct node* root){ // Base Case: An empty tree is complete Binary Tree if (root == NULL) return true; // Create an empty queue int rear, front; struct node **queue = createQueue(&front, &rear); // Create a flag variable which will be set true // when a non full node is seen bool flag = false; // Do level order traversal using queue. enQueue(queue, &rear, root); while(!isQueueEmpty(&front, &rear)) { struct node *temp_node = deQueue(queue, &front); /* Ceck if left child is present*/ if(temp_node->left) { // If we have seen a non full node, and we see a node // with non-empty left child, then the given tree is not // a complete Binary Tree if (flag == true) return false; enQueue(queue, &rear, temp_node->left); // Enqueue Left Child } else // If this a non-full node, set the flag as true flag = true; /* Ceck if right child is present*/ if(temp_node->right) { // If we have seen a non full node, and we see a node // with non-empty left child, then the given tree is not // a complete Binary Tree if(flag == true) return false; enQueue(queue, &rear, temp_node->right); // Enqueue Right Child } else // If this a non-full node, set the flag as true flag = true; } // If we reach here, then the tree is complete Bianry Tree return true;}Time Complexity is O(n)
Space Complexity is O(n)
public boolean checkIfComplete() { if (root == null) return true; LinkedList<TreeNode> queue = new LinkedList(); boolean nonFullNodeSeen = false; queue.add(root); while (!queue.isEmpty()) { TreeNode currentNode = queue.remove(); if ((currentNode.left != null) && (currentNode.right != null)) { // there should not be any non-leaf node after first non full-node is seen if (nonFullNodeSeen) { return false; } queue.add(currentNode.left); queue.add(currentNode.right); } if ((currentNode.left != null) && (currentNode.right == null)) { // there should not be any non-leaf node after first non full-node is seen if (nonFullNodeSeen) { return false; } // this is the first non-full node seen nonFullNodeSeen = true; queue.add(currentNode.left); } // this should never be the case for a complete binary tree if ((currentNode.left == null) && (currentNode.right != null)) { return false; } } return true; }http://www.geeksforgeeks.org/check-whether-binary-tree-complete-not-set-2-recursive-solution/
n the array representation of a binary tree, if the parent node is assigned an index of ‘i’ and left child gets assigned an index of ‘2*i + 1’ while the right child is assigned an index of ‘2*i + 2’. If we represent the binary tree below as an array with the respective indices assigned to the different nodes of the tree below are shown below:-
As can be seen from the above figure, the assigned indices in case of a complete binary tree will strictly less be than the number of nodes in the complete binary tree. Below is the example of non-complete binary tree with the assigned array indices. As can be seen the assigned indices are equal to the number of nodes in the binary tree. Hence this tree is not a complete binary tree.
Hence we proceed in the following manner in order to check if the binary tree is complete binary tree.
- Calculate the number of nodes (count) in the binary tree.
- Start recursion of the binary tree from the root node of the binary tree with index (i) being set as 0 and the number of nodes in the binary (count).
- If the current node under examination is NULL, then the tree is a complete binary tree. Return true.
- If index (i) of the current node is greater than or equal to the number of nodes in the binary tree (count) i.e. (i>= count), then the tree is not a complete binary. Return false.
- Recursively check the left and right sub-trees of the binary tree for same condition. For the left sub-tree use the index as (2*i + 1) while for the right sub-tree use the index as (2*i + 2).
The time complexity of the above algorithm is O(n).
unsigned int countNodes(struct Node* root){ if (root == NULL) return (0); return (1 + countNodes(root->left) + countNodes(root->right));}/* This function checks if the binary tree is complete or not */bool isComplete (struct Node* root, unsigned int index, unsigned int number_nodes){ // An empty tree is complete if (root == NULL) return (true); // If index assigned to current node is more than // number of nodes in tree, then tree is not complete if (index >= number_nodes) return (false); // Recur for left and right subtrees return (isComplete(root->left, 2*index + 1, number_nodes) && isComplete(root->right, 2*index + 2, number_nodes));}Read full article from Check whether a given Binary Tree is Complete or not | GeeksforGeeks
