Good Nodes

Easy
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Average time to solve is 15m
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MicrosoftOracleMorgan Stanley

Problem statement

You are given the root node of a binary tree consisting of ‘N’ nodes. Your task is to find the number of good nodes in the given binary tree.

A good node is defined as good if there are no nodes with a value greater than X’s value in the path from the root to X.

You are given a root node ‘ROOT’.Your task is to return the number of good nodes.

Example: 
Elements are in the level order form. The input consists of values of nodes separated by a single space in a single line. In case a node is null, we take -1 in its place.

For example, the input for the tree depicted in the below image would be :

Example

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :
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).
Note : 
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
Detailed explanation ( Input/output format, Notes, Images )
Input Format: 
The first line of the input contains an integer, 'T,’ denoting the number of test cases.

The first line of each test case contains the elements of the tree in the level order form separated by a single space. If any node does not have a left or right child, take -1 in its place. Refer to the example for further clarification.
Output Format: 
For each test case, print an integer representing the total number of good nodes.

Print the output of each test case in a separate line.
Note: 
You do not need to print anything. It has already been taken care of. Just implement the given function.
Constraints: 
1 <= T <= 10
1 <= N <= 10^6
1 <= nodeVal <=10^5

Time limit: 1 sec
Sample Input 1:
2
3 1 9 -1 -1 8 10 -1 -1 -1 -1
2 1 4 -1 -1 3 -1 -1 -1
Sample Output 1:
3
2
Explanation of sample input 1:
For the first test case,

Example

Here, Node with value 3 , node with value 9, and node with value 10 are good nodes as there is no nodes with value greater than Node’s value in the path from the root to the node.
Hence, the answer is 3.

For the second test case,

Example

Here, Node with value 2 and node with value 4 are good nodes as there is no nodes with value greater than Node’s value in the path from the root to the node.
Hence, the answer is 2.
Sample Input 2:
2
3 -1 4 -1 1 -1 -1 
4 -1 5 5 -1 2 -1 3 -1 -1 5 -1 -1
Sample Output 2:
4 
2
Hint

Keep the track of maximum number.

Approaches (2)
Using Recursion

This approach will create a recursive function REC(‘CUR’,’MX’) that will return the number of good nodes in the subtree with root as ‘CUR’ node and ‘MX’ is the maximum number we encountered in the path from the root to ‘CUR’ node. We will first check whether ‘CUR’ node is good or not and then recursively call this function for its left and right subtree to find the number of good nodes in the left and right subtree, respectively.


 

Algorithm:

  • Defining the REC(‘CUR’,’MX’)
    • If the ‘’CUR’ is an empty node, return 0.
    • Set  ‘ANS’ as 0.
    • If ‘MX’ is less than or equal to CUR’s value:
      • CUR node is a good node.
      • Set ‘ANS’ as ‘ANS’ +1.
    • Now, update ‘MX’ as the maximum of ‘MX’ and CUR’s value.
    • Set ‘ANS’ as ‘ANS’ + REC(left node of ‘CUR’,MX).
    • Set ‘ANS’ as ‘ANS’ + REC(right node of ‘CUR’,MX).
    • Return ‘ANS’.

 

  • Set ‘ANS’ as REC(‘ROOT’, the value of ‘ROOT’).
  • Return ‘ANS’.
Time Complexity

O(N), where N is the number of nodes in the tree.

 

In this approach, we traverse each node once, and checking whether the node is good or not. Hence, the overall time complexity is O(N).

Space Complexity

O(N), where N is the number of nodes in the tree.

 

We use O(N) space complexity for recursive stack during recursion calls in the worst case. The space complexity O(N) is also used for storing the preorder traversal. Hence, the overall space complexity is O(N).

Code Solution
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Good Nodes
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