Diameter of Binary Tree

class Solution {
    private int maxDiameter;
    public int diameterOfBinaryTree(TreeNode root) {
        maxDiameter = 0;
        calculateMaxPath(root);
        return maxDiameter;
    }
    private int calculateMaxPath(TreeNode node){
        if(node == null) return 0;
        int leftMax = calculateMaxPath(node.left);
        int rightMax = calculateMaxPath(node.right);

        maxDiameter = Math.max(maxDiameter, leftMax + rightMax);

        return Math.max(leftMax, rightMax) + 1;
    }
}

The provided solution in Java is designed to compute the diameter of a binary tree, where the diameter is defined as the longest path between any two nodes in the tree. This path does not necessarily pass through the root. It employs a recursive approach with a helper function to efficiently determine the maximum path from any node.

Here's an explanation of how the code functions:

• The Solution class contains a private variable maxDiameter which tracks the maximum length found during recursion.
• The method diameterOfBinaryTree(TreeNode root) initializes maxDiameter and starts the recursive helper method calculateMaxPath where the actual calculation is performed.
• The calculateMaxPath method computes recursively for each node:
  • Base case: If the node is null, returns 0.
  • Recursively finds the maximum path length for the left child leftMax and right child rightMax.
  • Updates the maxDiameter by comparing it to the sum of leftMax and rightMax. This step checks if connecting the two children through the current node results in a longer path than previously recorded.
  • Returns the height of the current subtree to its parent call, determined by the maximum height of its children plus one.

This approach optimizes the search by calculating diameter and subtree height in one pass, avoiding redundant traversals and ensuring efficiency. The method ultimately returns the value of maxDiameter after completing the recursive processing, which is the longest diameter of the binary tree.

class Solution:
    def treeDiameter(self, root: TreeNode) -> int:
        max_diameter = 0

        def max_path(node):
            if not node:
                return 0
            nonlocal max_diameter
            left_length = max_path(node.left)
            right_length = max_path(node.right)

            max_diameter = max(max_diameter, left_length + right_length)

            return max(left_length, right_length) + 1

        max_path(root)
        return max_diameter

This Python 3 solution aims to calculate the diameter of a binary tree. The diameter is defined as the longest path between any two nodes in the tree. This is calculated using a depth-first search (DFS) approach encapsulated within an inner function max_path.

• The solution class, Solution, has a method treeDiameter which takes root of type TreeNode as its argument.
• Inside the method, a local variable max_diameter is initialized to keep track of the maximum diameter found during traversal.
• The max_path inner function recursively computes the maximum path length from each node. It:
  • Returns 0 if the node is None, indicating the base case of recursion.
  • Uses a nonlocal declaration to modify max_diameter defined in the enclosing scope.
  • Computes the length of the paths from the current node to the left and right children.
  • Updates max_diameter to be the maximum value between the current max_diameter and the sum of left and right path lengths.
  • Returns the maximum path length from the node downward, adding one for the current node.
• Finally, the max_path function is called with the root node, and the maximum diameter accumulated during the traversal is returned.

This method effectively handles each node a single time, ensuring the function runs efficiently with time complexity loosely coupled with the number of nodes in the tree, thus providing a robust solution to determine the tree's diameter.