Java

/**
 Given a binary tree, you need to compute 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.

 Example:
 Given a binary tree
    1
   / \
  2   3
 / \
4   5
 Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

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

 */
public class Diameter_of_Binary_Tree {
    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {

        int max = 0;

        public int diameterOfBinaryTree(TreeNode root) {
            dfs(root);
            return max;
        }

        private int dfs(TreeNode root) {
            if (root == null) {
                return 0;
            }
            int leftDepth = dfs(root.left);
            int rightDepth = dfs(root.right);

            max = Math.max(max, leftDepth + rightDepth);

            return Math.max(leftDepth, rightDepth) + 1;
        }
    }
}

// OJ: https://leetcode.com/problems/diameter-of-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    int postorder(TreeNode *root) {
        if (!root) return 0;
        int left = postorder(root->left), right = postorder(root->right);
        ans = max(ans, left + right);
        return 1 + max(left, right);
    }
public:
    int diameterOfBinaryTree(TreeNode* root) {
        postorder(root);
        return ans;
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def diameterOfBinaryTree(self, root: TreeNode) -> int:
        def dfs(root):
            if root is None:
                return 0
            nonlocal ans
            left, right = dfs(root.left), dfs(root.right)
            ans = max(ans, left + right)
            return 1 + max(left, right)

        ans = 0
        dfs(root)
        return ans

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def diameterOfBinaryTree(self, root):
    """
    :type root: TreeNode
    :rtype: int
    """
    self.ans = 0

    def dfs(root):
      if not root:
        return 0
      left = dfs(root.left)
      right = dfs(root.right)
      self.ans = max(self.ans, left + right)
      return max(left, right) + 1

    dfs(root)
    return self.ans

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int path;

    public int diameterOfBinaryTree(TreeNode root) {
        path = 1;
        depthFirstSearch(root);
        return path - 1;        
    }

    public int depthFirstSearch(TreeNode node) {
        if (node == null)
            return 0;
        int leftDepth = depthFirstSearch(node.left);
        int rightDepth = depthFirstSearch(node.right);
        path = Math.max(path, leftDepth + rightDepth + 1);
        return Math.max(leftDepth, rightDepth) + 1;
    }
}

// OJ: https://leetcode.com/problems/diameter-of-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
    int ans = 0;
    int postorder(TreeNode *root) {
        if (!root) return 0;
        int left = postorder(root->left), right = postorder(root->right);
        ans = max(ans, left + right);
        return 1 + max(left, right);
    }
public:
    int diameterOfBinaryTree(TreeNode* root) {
        postorder(root);
        return ans;
    }
};

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def diameterOfBinaryTree(self, root: TreeNode) -> int:
        def dfs(root):
            if root is None:
                return 0
            nonlocal ans
            left, right = dfs(root.left), dfs(root.right)
            ans = max(ans, left + right)
            return 1 + max(left, right)

        ans = 0
        dfs(root)
        return ans

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def diameterOfBinaryTree(self, root):
    """
    :type root: TreeNode
    :rtype: int
    """
    self.ans = 0

    def dfs(root):
      if not root:
        return 0
      left = dfs(root.left)
      right = dfs(root.right)
      self.ans = max(self.ans, left + right)
      return max(left, right) + 1

    dfs(root)
    return self.ans

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