Height of Binary Tree

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Problem statement

The height of a tree is equal to the number of nodes on the longest path from the root to a leaf.


You are given an arbitrary binary tree consisting of 'n' nodes where each node is associated with a certain value.


Find out the height of the tree.


Example : 
Input: Let the binary tree be:


Output: 2

Explanation: The root node is 3, and the leaf nodes are 1 and 2.

There are two nodes visited when traversing from 3 to 1.
There are two nodes visited when traversing from 3 to 2.

Therefore the height of the binary tree is 2.
Detailed explanation ( Input/output format, Notes, Images )
Input Format : 
The first and only line contains the values of the tree’s nodes in the level order form ( -1 for NULL node). Refer to the example for further clarification.

Consider the binary tree:

altImage

The input of the tree depicted in the image above will be like: 
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).


Output Format : 
Print a single integer denoting the height of the given binary tree.


Note : 
You do not need to print anything; it has already been taken care of. Just implement the given function.
Sample Input 1:
3 1 2 -1 -1 -1 -1


Sample Output 1:
2


Explanation for sample input 1:
The given tree is:


The root node is 3, and the leaf nodes are 1 and 2.

There are two nodes visited when traversing from 3 to 1.
There are two nodes visited when traversing from 3 to 2.

Therefore the height of the binary tree is 2.


Sample Input 2:
3 -1 1 2 -1 -1 -1


Sample Output 2:
3


Explanation of sample input 2 :
The given tree is:


The root node is 3, and there is only one leaf node, which is 2.

All three nodes are visited while traversing from 3 to 2.

Therefore the height of the binary tree is 3.


Sample Input 3:
2 -1 -1


Sample Output 3:
1


Expected time complexity :
The expected time complexity is O(n).


Constraints :
1 <= 'n' <= 10000

Time Limit: 1 second
Hint

Tree/ Traversal Techniques( DFS )

Approaches (2)
DFS

To find the depth of the tree, We will do the DFS on the tree starting from the root node. 

In DFS, we visit all the nodes (child and grandchild nodes) of one child node before going to the second child node, which will help us determine the tree's height. We will take the maximum of both the child’s height.

We will go to the left node and increase one if it is not NULL and do DFS on the left node, the similar thing we will do on the correct node and return the max of the left and right node.

The steps are as follows:

heightOfBinaryTree(TreeNode 'root'):

  1. If 'root' is NULL:
    • Return 0, because it is not counted as a node.
  2. Else:
    • Initialize ‘depthLeft’ = heightOfBinaryTree(root.left ).
    • Initialize ‘depthRight’ = heightOfBinaryTree(root.right ).
    • We take the maximum of both the depth because height is the maximum number of edges encountered from the root to one or more leaf nodes.
    • If depthLeft > depthRight:
      • Return ‘depthLeft + 1’.
    • Else:
      • Return ‘depthRight + 1’.
    • We add the +1 on each node call to represent the edge connecting the current node to that child node.
Time Complexity

O(n), Where 'n' is the number of nodes in the tree.
We are traversing the whole tree, and the number of nodes in the tree is 'n'.

Hence time complexity is O(n).

Space Complexity

O(n), Where 'n' is the number of nodes in the tree.
The recursion stack space might consume O(n) space in the worst case.

Hence, the overall Space Complexity is O(n).

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