LeetCode 543: Diameter of Binary Tree

URL: https://leetcode.com/problems/diameter-of-binary-tree/description/

Problem Statement

Diameter: Longest path from any two nodes within a tree

Example Testcase

Input: root=[1,null,2,3,4,5]
Output:3`

Method

Intuition: Diameter = max(left subtree, right subtree) per node and DFS

Initial Pseudocode

findDiameter (node) {
	if (root == null) return;
	
    leftHeight = findLeftHeight(node->left)
    rightHeight = findRightHeight(node->right)
    
    currentDiameter = max(currentDiameter, leftHeight + rightHeight)
    
    # DFS portion
    findDiameter(node->left)
    findDiameter(node->right)

}

Time & Space Complexity

$O(N) \times O(W) \approx O(N^2)$

• not good, especially for skewed tree!

Optimization

Problem: Recalculating height multiple times
Objective: Reduce height calculation

New Pseudocode

int findDiameter(node, currentDiameter) {
    if (node == null) return 0;

    leftHeight = findDiameter(node->left, currentDiameter)
    rightHeight = findDiameter(node->right, currentDiameter)

    currentDiameter = max(currentDiameter, leftHeight + rightHeight)

    return max(leftHeight, rightHeight) + 1
}

C++

/**
 * 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 {
public:
    int diameterOfBinaryTree(TreeNode* root) {
        int res = 0;
        dfs(root, res);
        return res;
    }

    int dfs(TreeNode* root, int& res) {
        if (!root) {
            return 0;
        }

        int left = dfs(root->left, res);
        int right = dfs(root->right, res);

        res = max(res, left + right);
        return 1 + max(left, right);

    }
};

Time Complexity:

$O(N)$