# 543. Diameter of Binary Tree

## Description

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

## Solutions

Similar to problem 687. Longest Univalue Path.

• /**
* 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 ans;

public int diameterOfBinaryTree(TreeNode root) {
ans = 0;
dfs(root);
return ans;
}

private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int left = dfs(root.left);
int right = dfs(root.right);
ans = Math.max(ans, left + right);
return 1 + Math.max(left, right);
}
}

• /**
* 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 ans;

int diameterOfBinaryTree(TreeNode* root) {
ans = 0;
dfs(root);
return ans;
}

int dfs(TreeNode* root) {
if (!root) return 0;
int left = dfs(root->left);
int right = dfs(root->right);
ans = max(ans, left + right);
return 1 + max(left, right);
}
};

• # 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.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func diameterOfBinaryTree(root *TreeNode) int {
ans := 0
var dfs func(root *TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
left, right := dfs(root.Left), dfs(root.Right)
ans = max(ans, left+right)
return 1 + max(left, right)
}
dfs(root)
return ans
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function diameterOfBinaryTree(root: TreeNode | null): number {
let res = 0;
const dfs = (root: TreeNode | null) => {
if (root == null) {
return 0;
}
const { left, right } = root;
const l = dfs(left);
const r = dfs(right);
res = Math.max(res, l + r);
return Math.max(l, r) + 1;
};
dfs(root);
return res;
}


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut i32) -> i32 {
if root.is_none() {
return 0;
}
let root = root.as_ref().unwrap().as_ref().borrow();
let left = Self::dfs(&root.left, res);
let right = Self::dfs(&root.right, res);
*res = (*res).max(left + right);
left.max(right) + 1
}

pub fn diameter_of_binary_tree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut res = 0;
Self::dfs(&root, &mut res);
res
}
}