Problem
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 then
elements divided byn
and rounded down to the nearest integer.A subtree of
root
is a tree consisting ofroot
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
Solution
/**
* 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 averageOfSubtree = function(root) {
return helper(root).res;
};
var helper = function(root) {
if (!root) return { sum: 0, num: 0, res: 0 };
var left = helper(root.left);
var right = helper(root.right);
var sum = left.sum + right.sum + root.val;
var num = left.num + right.num + 1;
var res = left.res + right.res + (Math.floor(sum / num) === root.val ? 1 : 0);
return { sum, num, res };
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).