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二叉树
题目描述
给你一个二叉树的根结点 root
,请返回出现次数最多的子树元素和。如果有多个元素出现的次数相同,返回所有出现次数最多的子树元素和(不限顺序)。
一个结点的 「子树元素和」 定义为以该结点为根的二叉树上所有结点的元素之和(包括结点本身)。
示例 1:
输入: root = [5,2,-3]
输出: [2,-3,4]
示例 2:
输入: root = [5,2,-5]
输出: [2]
提示:
节点数在 [1, 104 ]
范围内
-105 <= Node.val <= 105
解法
方法一
Python3 Java C++ Go TypeScript Rust
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20 # 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 findFrequentTreeSum ( self , root : TreeNode ) -> List [ int ]:
def dfs ( root ):
if root is None :
return 0
left , right = dfs ( root . left ), dfs ( root . right )
s = root . val + left + right
counter [ s ] += 1
return s
counter = Counter ()
dfs ( root )
mx = max ( counter . values ())
return [ k for k , v in counter . items () if v == mx ]
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46 /**
* 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 Map < Integer , Integer > counter ;
private int mx ;
public int [] findFrequentTreeSum ( TreeNode root ) {
counter = new HashMap <> ();
mx = Integer . MIN_VALUE ;
dfs ( root );
List < Integer > res = new ArrayList <> ();
for ( Map . Entry < Integer , Integer > entry : counter . entrySet ()) {
if ( entry . getValue () == mx ) {
res . add ( entry . getKey ());
}
}
int [] ans = new int [ res . size () ] ;
for ( int i = 0 ; i < res . size (); ++ i ) {
ans [ i ] = res . get ( i );
}
return ans ;
}
private int dfs ( TreeNode root ) {
if ( root == null ) {
return 0 ;
}
int s = root . val + dfs ( root . left ) + dfs ( root . right );
counter . put ( s , counter . getOrDefault ( s , 0 ) + 1 );
mx = Math . max ( mx , counter . get ( s ));
return s ;
}
}
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34 /**
* 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 :
unordered_map < int , int > counter ;
int mx = 0 ;
vector < int > findFrequentTreeSum ( TreeNode * root ) {
mx = INT_MIN ;
dfs ( root );
vector < int > ans ;
for ( auto & entry : counter )
if ( entry . second == mx )
ans . push_back ( entry . first );
return ans ;
}
int dfs ( TreeNode * root ) {
if ( ! root ) return 0 ;
int s = root -> val + dfs ( root -> left ) + dfs ( root -> right );
++ counter [ s ];
mx = max ( mx , counter [ s ]);
return s ;
}
};
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32 /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func findFrequentTreeSum ( root * TreeNode ) [] int {
counter := make ( map [ int ] int )
mx := 0
var dfs func ( root * TreeNode ) int
dfs = func ( root * TreeNode ) int {
if root == nil {
return 0
}
s := root . Val + dfs ( root . Left ) + dfs ( root . Right )
counter [ s ] ++
if mx < counter [ s ] {
mx = counter [ s ]
}
return s
}
dfs ( root )
var ans [] int
for k , v := range counter {
if v == mx {
ans = append ( ans , k )
}
}
return ans
}
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36 /**
* 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 findFrequentTreeSum ( root : TreeNode | null ) : number [] {
const map = new Map < number , number > ();
let max = 0 ;
const dfs = ( root : TreeNode | null ) => {
if ( root == null ) {
return 0 ;
}
const { val , left , right } = root ;
const sum = val + dfs ( left ) + dfs ( right );
map . set ( sum , ( map . get ( sum ) ?? 0 ) + 1 );
max = Math . max ( max , map . get ( sum ));
return sum ;
};
dfs ( root );
const res = [];
for ( const [ k , v ] of map ) {
if ( v === max ) {
res . push ( k );
}
}
return res ;
}
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50 // 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 ;
use std :: collections :: HashMap ;
impl Solution {
fn dfs (
root : & Option < Rc < RefCell < TreeNode >>> ,
map : & mut HashMap < i32 , i32 > ,
max : & mut i32
) -> i32 {
if root . is_none () {
return 0 ;
}
let node = root . as_ref (). unwrap (). borrow ();
let sum = node . val + Self :: dfs ( & node . left , map , max ) + Self :: dfs ( & node . right , map , max );
map . insert ( sum , map . get ( & sum ). unwrap_or ( & 0 ) + 1 );
* max = ( * max ). max ( map [ & sum ]);
sum
}
pub fn find_frequent_tree_sum ( root : Option < Rc < RefCell < TreeNode >>> ) -> Vec < i32 > {
let mut map = HashMap :: new ();
let mut max = 0 ;
let mut res = Vec :: new ();
Self :: dfs ( & root , & mut map , & mut max );
for ( k , v ) in map . into_iter () {
if v == max {
res . push ( k );
}
}
res
}
}
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