LeetCode //C - 222. Count Complete Tree Nodes

222. Count Complete Tree Nodes

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 $2^h$ 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 [$0,5\ast 10^4$].
• $0<=Node.val<=5\ast 10^4$
• The tree is guaranteed to be complete.

From: LeetCode
Link: 222. Count Complete Tree Nodes

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Solution:

Ideas：

Main Idea:
The main idea is to utilize a combination of the properties of a binary tree and binary search. Given that the tree is complete, we can determine the depth (or height) of the tree and then perform a binary search to find the number of nodes in the last level.

Code Explanation:

1. depth function:

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• Computes the depth (or height) of the tree by simply traversing down the leftmost path until we hit a null node. This takes $O(h)$ time where h is the height of the tree.

2. exists function:

• This function checks if a node exists at a given index on the last level of the tree.
• The function uses binary representation of the index to traverse from root to the desired node. For each bit in the binary representation:
  • If the bit is 0, move to the left child.
  • If the bit is 1, move to the right child.
• At the end of this traversal, if the node exists, the function returns true, otherwise false.

3. countNodes function:

• First, if the tree is empty, it immediately returns 0.
• It then computes the depth of the tree.
• If the depth is 1 (only root exists), it returns 1.
• For trees with depth greater than 1, it performs a binary search to find the rightmost node’s index on the last level:
  • The search range for the index is from 0 to $2^h-1$, where h is the depth of the tree minus 1 (since we’re only interested in the last level).
  • Using the exists function, it checks the existence of a node for a given index.
  • If the node exists for the current pivot index, it adjusts the search range to the right half, otherwise, it adjusts to the left half.
• Once the binary search completes, the left pointer indicates the total number of nodes in the last level.
• The total count of nodes in the tree is the sum of nodes in the first $h-1$ levels (which is $2^{h-1}-1$) and the number of nodes in the last level.

Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int depth(struct TreeNode* root) {
    
    
    int d = 0;
    while (root) {
    
    
        root = root->left;
        d++;
    }
    return d;
}

bool exists(int idx, int d, struct TreeNode* root) {
    
    
    int left = 0, right = (1 << d) - 1;
    for (int i = 0; i < d; i++) {
    
    
        int pivot = left + (right - left) / 2;
        if (idx <= pivot) {
    
    
            root = root->left;
            right = pivot;
        } else {
    
    
            root = root->right;
            left = pivot + 1;
        }
    }
    return root != NULL;
}

int countNodes(struct TreeNode* root) {
    
    
    if (!root) return 0;
    
    int d = depth(root) - 1;
    if (d == 0) return 1;
    
    int left = 0, right = (1 << d) - 1;
    while (left <= right) {
    
    
        int pivot = left + (right - left) / 2;
        if (exists(pivot, d, root)) {
    
    
            left = pivot + 1; // Adjust here. Move to the next possible node.
        } else {
    
    
            right = pivot - 1;
        }
    }
    
    return (1 << d) - 1 + left; // left now represents the count of nodes in the last level.
}