Tree
Depth-First Search
Binary Tree
Description
You are given the root
of a binary tree and an integer distance
. A pair of two different leaf nodes of a binary tree is said to be good if the length of the shortest path between them is less than or equal to distance
.
Return the number of good leaf node pairs in the tree.
Example 1:
Input: root = [1,2,3,null,4], distance = 3
Output: 1
Explanation: The leaf nodes of the tree are 3 and 4 and the length of the shortest path between them is 3. This is the only good pair.
Example 2:
Input: root = [1,2,3,4,5,6,7], distance = 3
Output: 2
Explanation: The good pairs are [4,5] and [6,7] with shortest path = 2. The pair [4,6] is not good because the length of ther shortest path between them is 4.
Example 3:
Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3
Output: 1
Explanation: The only good pair is [2,5].
Constraints:
The number of nodes in the tree
is in the range [1, 210 ].
1 <= Node.val <= 100
1 <= distance <= 10
Solutions
Solution 1
Python3 Java C++ Go TypeScript JavaScript
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32 # 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 countPairs ( self , root : TreeNode , distance : int ) -> int :
def dfs ( root , cnt , i ):
if root is None or i >= distance :
return
if root . left is None and root . right is None :
cnt [ i ] += 1
return
dfs ( root . left , cnt , i + 1 )
dfs ( root . right , cnt , i + 1 )
if root is None :
return 0
ans = self . countPairs ( root . left , distance ) + self . countPairs (
root . right , distance
)
cnt1 = Counter ()
cnt2 = Counter ()
dfs ( root . left , cnt1 , 1 )
dfs ( root . right , cnt2 , 1 )
for k1 , v1 in cnt1 . items ():
for k2 , v2 in cnt2 . items ():
if k1 + k2 <= distance :
ans += v1 * v2
return ans
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47 /**
* 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 countPairs ( TreeNode root , int distance ) {
if ( root == null ) {
return 0 ;
}
int ans = countPairs ( root . left , distance ) + countPairs ( root . right , distance );
int [] cnt1 = new int [ distance ] ;
int [] cnt2 = new int [ distance ] ;
dfs ( root . left , cnt1 , 1 );
dfs ( root . right , cnt2 , 1 );
for ( int i = 0 ; i < distance ; ++ i ) {
for ( int j = 0 ; j < distance ; ++ j ) {
if ( i + j <= distance ) {
ans += cnt1 [ i ] * cnt2 [ j ] ;
}
}
}
return ans ;
}
void dfs ( TreeNode root , int [] cnt , int i ) {
if ( root == null || i >= cnt . length ) {
return ;
}
if ( root . left == null && root . right == null ) {
++ cnt [ i ] ;
return ;
}
dfs ( root . left , cnt , i + 1 );
dfs ( root . right , cnt , i + 1 );
}
}
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40 /**
* 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 countPairs ( TreeNode * root , int distance ) {
if ( ! root ) return 0 ;
int ans = countPairs ( root -> left , distance ) + countPairs ( root -> right , distance );
vector < int > cnt1 ( distance );
vector < int > cnt2 ( distance );
dfs ( root -> left , cnt1 , 1 );
dfs ( root -> right , cnt2 , 1 );
for ( int i = 0 ; i < distance ; ++ i ) {
for ( int j = 0 ; j < distance ; ++ j ) {
if ( i + j <= distance ) {
ans += cnt1 [ i ] * cnt2 [ j ];
}
}
}
return ans ;
}
void dfs ( TreeNode * root , vector < int >& cnt , int i ) {
if ( ! root || i >= cnt . size ()) return ;
if ( ! root -> left && ! root -> right ) {
++ cnt [ i ];
return ;
}
dfs ( root -> left , cnt , i + 1 );
dfs ( root -> right , cnt , i + 1 );
}
};
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38 /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func countPairs ( root * TreeNode , distance int ) int {
if root == nil {
return 0
}
ans := countPairs ( root . Left , distance ) + countPairs ( root . Right , distance )
cnt1 := make ([] int , distance )
cnt2 := make ([] int , distance )
dfs ( root . Left , cnt1 , 1 )
dfs ( root . Right , cnt2 , 1 )
for i , v1 := range cnt1 {
for j , v2 := range cnt2 {
if i + j <= distance {
ans += v1 * v2
}
}
}
return ans
}
func dfs ( root * TreeNode , cnt [] int , i int ) {
if root == nil || i >= len ( cnt ) {
return
}
if root . Left == nil && root . Right == nil {
cnt [ i ] ++
return
}
dfs ( root . Left , cnt , i + 1 )
dfs ( root . Right , cnt , i + 1 )
}
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32 function countPairs ( root : TreeNode | null , distance : number ) : number {
const pairs : number [][] = [];
const dfs = ( node : TreeNode | null ) : number [][] => {
if ( ! node ) return [];
if ( ! node . left && ! node . right ) return [[ node . val , 1 ]];
const left = dfs ( node . left );
const right = dfs ( node . right );
for ( const [ x , dx ] of left ) {
for ( const [ y , dy ] of right ) {
if ( dx + dy <= distance ) {
pairs . push ([ x , y ]);
}
}
}
const res : number [][] = [];
for ( const arr of [ left , right ]) {
for ( const x of arr ) {
if ( ++ x [ 1 ] <= distance ) res . push ( x );
}
}
return res ;
};
dfs ( root );
return pairs . length ;
}
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45 /**
* 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
* @param {number} distance
* @return {number}
*/
var countPairs = function ( root , distance ) {
const pairs = [];
const dfs = node => {
if ( ! node ) return [];
if ( ! node . left && ! node . right ) return [[ node . val , 1 ]];
const left = dfs ( node . left );
const right = dfs ( node . right );
for ( const [ x , dx ] of left ) {
for ( const [ y , dy ] of right ) {
if ( dx + dy <= distance ) {
pairs . push ([ x , y ]);
}
}
}
const res = [];
for ( const arr of [ left , right ]) {
for ( const x of arr ) {
if ( ++ x [ 1 ] <= distance ) res . push ( x );
}
}
return res ;
};
dfs ( root );
return pairs . length ;
};
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