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
Solution 1: Enumeration + DFS
We can enumerate each node of the binary tree, and for each node, calculate the maximum depth of its left and right subtrees, l and r, respectively. The diameter of the node is l + r. The maximum diameter among all nodes is the diameter of the binary tree.
The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes in the binary tree.
PythonJavaC++GoTypeScriptRustJavaScriptC#C
# 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(code-box)
/**
* 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) {
dfs(root);
return ans;
}
private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int l = dfs(root.left);
int r = dfs(root.right);
ans = Math.max(ans, l + r);
return 1 + Math.max(l, r);
}
}(code-box)
/**
* 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 ans = 0;
auto dfs = [&](this auto&& dfs, TreeNode* root) -> int {
if (!root) {
return 0;
}
int l = dfs(root->left);
int r = dfs(root->right);
ans = max(ans, l + r);
return 1 + max(l, r);
};
dfs(root);
return ans;
}
};(code-box)
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func diameterOfBinaryTree(root *TreeNode) (ans int) {
var dfs func(root *TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
l, r := dfs(root.Left), dfs(root.Right)
ans = max(ans, l+r)
return 1 + max(l, r)
}
dfs(root)
return
}(code-box)
/**
* 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 ans = 0;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
const [l, r] = [dfs(root.left), dfs(root.right)];
ans = Math.max(ans, l + r);
return 1 + Math.max(l, r);
};
dfs(root);
return ans;
}(code-box)
// 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::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn diameter_of_binary_tree(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut ans = 0;
fn dfs(root: Option<Rc<RefCell<TreeNode>>>, ans: &mut i32) -> i32 {
match root {
Some(node) => {
let node = node.borrow();
let l = dfs(node.left.clone(), ans);
let r = dfs(node.right.clone(), ans);
*ans = (*ans).max(l + r);
1 + l.max(r)
}
None => 0,
}
}
dfs(root, &mut ans);
ans
}
}(code-box)
/**
* 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 ans = 0;
const dfs = root => {
if (!root) {
return 0;
}
const [l, r] = [dfs(root.left), dfs(root.right)];
ans = Math.max(ans, l + r);
return 1 + Math.max(l, r);
};
dfs(root);
return ans;
};(code-box)
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
private int ans;
public int DiameterOfBinaryTree(TreeNode root) {
dfs(root);
return ans;
}
private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int l = dfs(root.left);
int r = dfs(root.right);
ans = Math.Max(ans, l + r);
return 1 + Math.Max(l, r);
}
}(code-box)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int dfs(struct TreeNode* root, int* ans) {
if (root == NULL) {
return 0;
}
int l = dfs(root->left, ans);
int r = dfs(root->right, ans);
if (l + r > *ans) {
*ans = l + r;
}
return 1 + (l > r ? l : r);
}
int diameterOfBinaryTree(struct TreeNode* root) {
int ans = 0;
dfs(root, &ans);
return ans;
}(code-box)
