543. Diameter of Binary Tree

📝 Problem Details

• Title: 543. Diameter of Binary Tree
• Link: https://leetcode.com/problems/diameter-of-binary-tree/
• Difficulty: Easy
• Tags/Categories: Trees

💭What Were My Initial Thoughts?

similar to maximum depth of a binary tree, except we are essentially finding the max depths for the right and left subtrees of the root and adding them together

🤔What Did I Struggle With?

- when to use the provided function for DFS and when to create a helper function
	- depends on the persistence of the result

💡 Explanation of Solution

- dfs helper function that takes in the TreeNode and a reference to the result
	- recursively explore the left and right subtrees
	- calculate the res by taking the max between the current res and the sum between the left and right subtree recursive values
	- return 1 + max(left,right)

⌛ Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(log n) - O(n) due to recursion stack depending on balance of tree

💻 Implementation of Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    int dfs(TreeNode* root, int& res) {
        if(root == nullptr) return 0;
 
        int leftSubtree = dfs(root->left, res);
        int rightSubtree = dfs(root->right, res);
        res = max(res, leftSubtree + rightSubtree);
        return 1 + max(leftSubtree, rightSubtree);
    }
 
public:
    int diameterOfBinaryTree(TreeNode* root) {
        int res = 0;
        dfs(root, res);
        return res;
    }
};