Medium
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:
n
elements is the sum of the n
elements divided by n
and rounded down to the nearest integer.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:
[1, 1000]
.0 <= Node.val <= 1000
import com_github_leetcode.TreeNode;
public class Solution {
private int ans = 0;
public int averageOfSubtree(TreeNode root) {
dfs(root);
return ans;
}
private int[] dfs(TreeNode node) {
if (node == null) {
return new int[] {0, 0};
}
int[] left = dfs(node.left);
int[] right = dfs(node.right);
int currsum = left[0] + right[0] + node.val;
int currcount = left[1] + right[1] + 1;
if (currsum / currcount == node.val) {
++ans;
}
return new int[] {currsum, currcount};
}
}