222. Count Complete Tree Nodes

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222. Count Complete Tree Nodes

Description

Difficulty: Easy

Related Topics: Binary Search, Tree, Depth-First Search, Binary Tree

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:

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Input: root = [1,2,3,4,5,6]
Output: 6

Example 2:

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Input: root = []
Output: 0

Example 3:

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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.

Hints/Notes

  • Check if it’s a full BST to improve the time complexity

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int countNodes(TreeNode* root) {
        int count = 0;
        if (!root) return count;
        int leftDepth = 0, rightDepth = 0;
        TreeNode* tmp = root;
        while (tmp) {
            leftDepth++;
            tmp = tmp->left;
        }
        tmp = root;
        while (tmp) {
            rightDepth++;
            tmp = tmp->right;
        }
        if (leftDepth == rightDepth) {
            return pow(2, leftDepth) - 1;
        }
        int leftCount = countNodes(root->left);
        int rightCount = countNodes(root->right);
        return leftCount + rightCount + 1;
    }
};