Tree
Depth-First Search
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
Solution 1
Python3 Java C++ Go TypeScript Rust C
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19 # 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
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34 /**
* 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 );
}
}
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29 /**
* 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 );
}
};
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22 /**
* 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
}
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29 /**
* 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 ;
}
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38 // 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 {
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
}
}
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26 /**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
#define max(a, b) (((a) > (b)) ? (a) : (b))
int dfs ( struct TreeNode * root , int * res ) {
if ( ! root ) {
return 0 ;
}
int left = dfs ( root -> left , res );
int right = dfs ( root -> right , res );
* res = max ( * res , left + right );
return max ( left , right ) + 1 ;
}
int diameterOfBinaryTree ( struct TreeNode * root ) {
int res = 0 ;
dfs ( root , & res );
return res ;
}
Solution 2
Python3
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41 # 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 build ( root ):
if root is None :
return
nonlocal d
if root . left :
d [ root ] . add ( root . left )
d [ root . left ] . add ( root )
if root . right :
d [ root ] . add ( root . right )
d [ root . right ] . add ( root )
build ( root . left )
build ( root . right )
def dfs ( u , t ):
nonlocal ans , vis , d , next
if u in vis :
return
vis . add ( u )
if t > ans :
ans = t
next = u
for v in d [ u ]:
dfs ( v , t + 1 )
d = defaultdict ( set )
ans = 0
next = root
build ( root )
vis = set ()
dfs ( next , 0 )
vis . clear ()
dfs ( next , 0 )
return ans
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