Description# Given the root
of a binary tree, return its maximum depth .
A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: 3
Example 2:
Input: root = [1,null,2]
Output: 2
Constraints:
The number of nodes in the tree is in the range [0, 104 ]
. -100 <= Node.val <= 100
Solutions# Solution 1: Recursion# Recursively traverse the left and right subtrees, calculate the maximum depth of the left and right subtrees, and then take the maximum value plus 1 1 1 .
The time complexity is O ( n ) O(n) O ( n ) , where n n n is the number of nodes in the binary tree. Each node is traversed only once in the recursion.
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# 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 maxDepth ( self , root : TreeNode ) -> int :
if root is None :
return 0
l , r = self . maxDepth ( root . left ), self . maxDepth ( root . right )
return 1 + max ( l , r )
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/**
* 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 {
public int maxDepth ( TreeNode root ) {
if ( root == null ) {
return 0 ;
}
int l = maxDepth ( root . left );
int r = maxDepth ( root . right );
return 1 + Math . max ( l , r );
}
}
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/**
* 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 maxDepth ( TreeNode * root ) {
if ( ! root ) return 0 ;
int l = maxDepth ( root -> left ), r = maxDepth ( root -> right );
return 1 + max ( l , r );
}
};
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/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func maxDepth ( root * TreeNode ) int {
if root == nil {
return 0
}
l , r := maxDepth ( root . Left ), maxDepth ( root . Right )
return 1 + max ( l , r )
}
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/**
* 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 maxDepth ( root : TreeNode | null ) : number {
if ( root === null ) {
return 0 ;
}
return 1 + Math . max ( maxDepth ( root . left ), maxDepth ( root . right ));
}
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// 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 >>> ) -> i32 {
if root . is_none () {
return 0 ;
}
let node = root . as_ref (). unwrap (). borrow ();
1 + Self ::dfs ( & node . left ). max ( Self ::dfs ( & node . right ))
}
pub fn max_depth ( root : Option < Rc < RefCell < TreeNode >>> ) -> i32 {
Self ::dfs ( & root )
}
}
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/**
* 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 maxDepth = function ( root ) {
if ( ! root ) return 0 ;
const l = maxDepth ( root . left );
const r = maxDepth ( root . right );
return 1 + Math . max ( l , r );
};
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/**
* 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 maxDepth ( struct TreeNode * root ) {
if ( ! root ) {
return 0 ;
}
int left = maxDepth ( root -> left );
int right = maxDepth ( root -> right );
return 1 + max ( left , right );
}
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