Given the root of a binary tree, return the number of nodes where the value of the node is equal to the average of the values in its subtree.
Note:
The average of n elements is the sum of the n elements divided by n and rounded down to the nearest integer.
A subtree of root is a tree consisting of root and all of its descendants.
Example 1:
Input: root = [4,8,5,0,1,null,6]
Output: 5
Explanation:
For the node with value 4: The average of its subtree is (4 + 8 + 5 + 0 + 1 + 6) / 6 = 24 / 6 = 4.
For the node with value 5: The average of its subtree is (5 + 6) / 2 = 11 / 2 = 5.
For the node with value 0: The average of its subtree is 0 / 1 = 0.
For the node with value 1: The average of its subtree is 1 / 1 = 1.
For the node with value 6: The average of its subtree is 6 / 1 = 6.
Example 2:
Input: root = [1]
Output: 1
Explanation: For the node with value 1: The average of its subtree is 1 / 1 = 1.
Constraints:
The number of nodes in the tree is in the range [1, 1000].
0 <= Node.val <= 1000
Solutions
Solution 1
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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 = rightclassSolution:defaverageOfSubtree(self,root:Optional[TreeNode])->int:defdfs(root):ifrootisNone:return0,0ls,ln=dfs(root.left)rs,rn=dfs(root.right)s=ls+rs+root.valn=ln+rn+1ifs//n==root.val:nonlocalansans+=1returns,nans=0dfs(root)returnans
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */funcaverageOfSubtree(root*TreeNode)int{ans:=0vardfsfunc(*TreeNode)(int,int)dfs=func(root*TreeNode)(int,int){ifroot==nil{return0,0}ls,ln:=dfs(root.Left)rs,rn:=dfs(root.Right)s:=ls+rs+root.Valn:=ln+rn+1ifs/n==root.Val{ans++}returns,n}dfs(root)returnans}