222. Count Complete Tree Nodes

Problem Description

Given the root of a complete binary tree, return the number of 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 2^h 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 * 10^4]
• 0 <= Node.val <= 5 * 10^4
• The tree is guaranteed to be complete

Python Solution

class Solution:
    def countNodes(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
            
        # Get the height of left and right subtree
        left_height = self.getLeftHeight(root)
        right_height = self.getRightHeight(root)
        
        # If heights are equal, use formula 2^h - 1
        if left_height == right_height:
            return (1 << left_height) - 1
            
        # Otherwise, recursively count nodes
        return 1 + self.countNodes(root.left) + self.countNodes(root.right)
        
    def getLeftHeight(self, node):
        height = 0
        while node:
            height += 1
            node = node.left
        return height
        
    def getRightHeight(self, node):
        height = 0
        while node:
            height += 1
            node = node.right
        return height

Java Solution

class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        // Get the height of left and right subtree
        int leftHeight = getLeftHeight(root);
        int rightHeight = getRightHeight(root);
        
        // If heights are equal, use formula 2^h - 1
        if (leftHeight == rightHeight) {
            return (1 << leftHeight) - 1;
        }
        
        // Otherwise, recursively count nodes
        return 1 + countNodes(root.left) + countNodes(root.right);
    }
    
    private int getLeftHeight(TreeNode node) {
        int height = 0;
        while (node != null) {
            height++;
            node = node.left;
        }
        return height;
    }
    
    private int getRightHeight(TreeNode node) {
        int height = 0;
        while (node != null) {
            height++;
            node = node.right;
        }
        return height;
    }
}

C++ Solution

class Solution {
public:
    int countNodes(TreeNode* root) {
        if (!root) {
            return 0;
        }
        
        // Get the height of left and right subtree
        int leftHeight = getLeftHeight(root);
        int rightHeight = getRightHeight(root);
        
        // If heights are equal, use formula 2^h - 1
        if (leftHeight == rightHeight) {
            return (1 << leftHeight) - 1;
        }
        
        // Otherwise, recursively count nodes
        return 1 + countNodes(root->left) + countNodes(root->right);
    }
    
private:
    int getLeftHeight(TreeNode* node) {
        int height = 0;
        while (node) {
            height++;
            node = node->left;
        }
        return height;
    }
    
    int getRightHeight(TreeNode* node) {
        int height = 0;
        while (node) {
            height++;
            node = node->right;
        }
        return height;
    }
};

JavaScript 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 countNodes = function(root) {
    if (!root) {
        return 0;
    }
    
    // Get the height of left and right subtree
    const leftHeight = getLeftHeight(root);
    const rightHeight = getRightHeight(root);
    
    // If heights are equal, use formula 2^h - 1
    if (leftHeight === rightHeight) {
        return (1 << leftHeight) - 1;
    }
    
    // Otherwise, recursively count nodes
    return 1 + countNodes(root.left) + countNodes(root.right);
};

function getLeftHeight(node) {
    let height = 0;
    while (node) {
        height++;
        node = node.left;
    }
    return height;
}

function getRightHeight(node) {
    let height = 0;
    while (node) {
        height++;
        node = node.right;
    }
    return height;
}

C# Solution

public class Solution {
    public int CountNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        
        // Get the height of left and right subtree
        int leftHeight = GetLeftHeight(root);
        int rightHeight = GetRightHeight(root);
        
        // If heights are equal, use formula 2^h - 1
        if (leftHeight == rightHeight) {
            return (1 << leftHeight) - 1;
        }
        
        // Otherwise, recursively count nodes
        return 1 + CountNodes(root.left) + CountNodes(root.right);
    }
    
    private int GetLeftHeight(TreeNode node) {
        int height = 0;
        while (node != null) {
            height++;
            node = node.left;
        }
        return height;
    }
    
    private int GetRightHeight(TreeNode node) {
        int height = 0;
        while (node != null) {
            height++;
            node = node.right;
        }
        return height;
    }
}