543. Diameter of Binary Tree

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, 104].
• -100 <= Node.val <= 100

Solution:

/**
 * 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 {
    int result = 0;
    public int diameterOfBinaryTree(TreeNode root) {
        if (root == null){
            return 0;
        }

        helper(root);

        return result;

    }

    private int helper(TreeNode root){
        if (root == null){
            return 0;
        }

        int left = helper(root.left);
        int right = helper(root.right);

        int cur = left + right; // not add 1 here because the diameter depends on the edge

        result = Math.max(cur, result);

        return Math.max(left, right) + 1;
    }
}

// TC: O(n)
// SC: O(n)

class Solution {
    private int diameter;
    public int diameterOfBinaryTree(TreeNode root) {
        diameter = 0;
        longestPath(root);
        return diameter;
    }
    private int longestPath(TreeNode node){
        if(node == null) return 0;
        // recursively find the longest path in
        // both left child and right child
        int leftPath = longestPath(node.left);
        int rightPath = longestPath(node.right);

        // update the diameter if left_path plus right_path is larger
        diameter = Math.max(diameter, leftPath + rightPath);

        // return the longest one between left_path and right_path;
        // remember to add 1 for the path connecting the node and its parent
        return Math.max(leftPath, rightPath) + 1;
    }
}