Medium
Given the root
of a complete binary tree, return the number of the nodes in the tree.
According to Wikipedia, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between 1
and 2h
nodes inclusive at the last level h
.
Design an algorithm that runs in less than O(n)
time complexity.
Example 1:
Input: root = [1,2,3,4,5,6]
Output: 6
Example 2:
Input: root = []
Output: 0
Example 3:
Input: root = [1]
Output: 1
Constraints:
[0, 5 * 104]
.0 <= Node.val <= 5 * 104
import com_github_leetcode.TreeNode
/*
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Solution {
fun countNodes(root: TreeNode?): Int {
if (root == null) {
return 0
}
val leftHeight = leftHeight(root)
val rightHeight = rightHeight(root)
// case 1: When Height(Left sub-tree) = Height(right sub-tree) 2^h - 1
return if (leftHeight == rightHeight) {
(1 shl leftHeight) - 1
} else {
1 + countNodes(root.left) + countNodes(root.right)
}
}
private fun leftHeight(root: TreeNode?): Int {
return if (root == null) {
0
} else {
1 + leftHeight(root.left)
}
}
private fun rightHeight(root: TreeNode?): Int {
return if (root == null) {
0
} else {
1 + rightHeight(root.right)
}
}
}