LeetCode-in-Java
222. Count Complete Tree Nodes
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:
- The number of nodes in the tree is in the range
[0, 5 * 104]. 0 <= Node.val <= 5 * 104- The tree is guaranteed to be complete.
Solution
import com_github_leetcode.TreeNode;
/*
* 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;
* }
* }
*/
public class Solution {
public int countNodes(TreeNode root) {
if (root == null) {
return 0;
}
int leftHeight = leftHeight(root);
int rightHeight = rightHeight(root);
// case 1: When Height(Left sub-tree) = Height(right sub-tree) 2^h - 1
if (leftHeight == rightHeight) {
return (1 << leftHeight) - 1;
} else {
return 1 + countNodes(root.left) + countNodes(root.right);
}
}
private int leftHeight(TreeNode root) {
if (root == null) {
return 0;
}
return 1 + leftHeight(root.left);
}
private int rightHeight(TreeNode root) {
if (root == null) {
return 0;
}
return 1 + rightHeight(root.right);
}
}