543. Diameter of Binary Tree

Problem

Given the root of a binary tree, return **the length of the *diameter* of the tree**.

The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

The length of a path between two nodes is represented by the number of edges between them.

  Example 1:

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].

Example 2:

Input: root = [1,2]
Output: 1

  Constraints:

• The number of nodes in the tree is in the range [1, 10^4].

• -100 <= Node.val <= 100

Solution (Java)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int diameter;

    public int diameterOfBinaryTree(TreeNode root) {
        diameter = 0;
        diameterOfBinaryTreeUtil(root);
        return diameter;
    }

    private int diameterOfBinaryTreeUtil(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftLength = root.left != null ? 1 + diameterOfBinaryTreeUtil(root.left) : 0;
        int rightLength = root.right != null ? 1 + diameterOfBinaryTreeUtil(root.right) : 0;
        diameter = Math.max(diameter, leftLength + rightLength);
        return Math.max(leftLength, rightLength);
    }
}

Solution (Javascript)

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var diameterOfBinaryTree = function(root) {
    let diameter = 0;
    dfs(root);
    return diameter;

    function dfs(node) {
        if (!node) return 0;

        const left = dfs(node.left);
        const right = dfs(node.right);

        // update diameter at every node
        diameter = Math.max(diameter, left + right);

        // update the max path of the current node
        // 1 is the height of the node itself + longest path of children
        return 1 + Math.max(left, right);
    }
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).