For the binary trees in the image below.
The left tree in the image is not a complete binary tree that’s why it is invalid and the right tree in the image is a valid complete binary tree which contains total 6 nodes.
The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases are as follows.
The only line of each test case contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. So -1 would not be a part of the tree nodes.
For example, the input for the tree depicted in the below image will be:
For example taking a tree:
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null(-1).
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
For each test case, print the number of nodes in the given binary tree.
Print the output of each test case in a separate line.
You don’t need to print anything, it has already been taken care of.
1 <= T <= 100
0 <= N <= 3000
0 <= val <= 10^5
Where ‘T’ is the number of test cases, and ‘N’ is the total number of nodes in the binary tree, and “val” is the value of the binary tree node
Time Limit: 1 sec
Our intuition is to calculate the number of nodes step by step from the leaf nodes. For that, we will use a recursive approach where we keep track of both the number of nodes in left and right subtrees.
For each subtree the total number of nodes will be :
1 + countNodes(root->left) + countNodes(root->right)
As we know the property of a complete binary tree, only the last level is supposed to be either totally or partially filled. Otherwise, rest all the levels are completely filled. So we do not need to iterate through the whole level which is fully filled, for that we can just keep a track of its height.
Steps are as follows :
Preorder Traversal
Inorder Traversal
Postorder Traversal
Height of Binary Tree
Locked Binary Tree