Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
O(n)
time complexity and O(1)
excluding the recursion call stack.HAPPY CASE
Input: root = [1,2,3,4,5]
Output: 3
Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
Input: root = [1,2]
Output: 1
EDGE CASE
Input: root = [1]
Output: 0
Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
If you are dealing with Binary Trees some common techniques you can employ to help you solve the problem:
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Recursively ask for child height in tree and add one for parent.
1. Create function to return the height of tree and collect diameter at each node
a. Basecase is a Null Node, return 0
b. Collect information regarding diameter of node
c. Recursively return the max between height of left node and right node and add one for current node.
2. Create variable to hold diameter information
3. Call the function to check diameter of each node
4. Return the diameter
⚠️ Common Mistakes
Implement the code to solve the algorithm.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
# Create function to return the height of tree and collect diameter at each node
def getDiameterOfBinaryTree(root: Optional[TreeNode]) -> int:
nonlocal diameter
# Basecase is a Null Node, return 0
if not root:
return 0
# Collect information regarding diameter of node
leftHeight = getDiameterOfBinaryTree(root.left)
rightHeight = getDiameterOfBinaryTree(root.right)
diameter = max(diameter, leftHeight + rightHeight)
# Recursively return the max between height of left node and right node and add one for current node
return max(leftHeight, rightHeight) + 1
# Create variable to hold diameter information
diameter = 0
# Call the function to check diameter of each node
getDiameterOfBinaryTree(root)
# Return the diameter
return diameter
class Solution
{
// Create variable to hold diameter information
int max;
// Create function to return the height of tree and collect diameter at each node
public int getDiameter(TreeNode root)
{
// Basecase is a Null Node, return 0
int h= 0;
if(root == null)
return 0;
// Collect information regarding diameter of node
int l= getDiameter(root.left);
int r= getDiameter(root.right);
// Recursively return the max between height of left node and right node and add one for current node
h= Math.max(l,r);//
max= Math.max(max,l+r+1);//updating the maximum diameter
h=h+1;//increasing heigh covering every node
return h; //returning the maximum height
}
public int diameterOfBinaryTree(TreeNode root)
{
max= Integer.MIN_VALUE;
// Call the function to check diameter of each node
getDiameter(root);
// Return the diameter
return max-1; //node=edge+1, i.e; edge=node-1
}
}
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Assume N
represents the number of nodes in binary tree
O(N)
, because we need to visit each node in the binary treeO(1)
because we exclude the recursive call stack. The recursive call stack may cost O(N)
we need to account for unbalanced tree. In the general case O(logN)
for balanced tree.