222. Count Complete Tree Nodes
1. Description
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.
2. Example
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
3. 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.
4. Solutions
Binary Search
n is the number of nodes in root
Time complexity: O($log^2n$)
Space complexity: O(1)
class Solution {
public:
int countNodes(TreeNode *root) {
if (root == nullptr) {
return 0;
}
int height = 0;
for (auto iter = root; iter != nullptr; iter = iter->left) {
++height;
}
if (height == 1) {
return 1;
}
int last_layer_count = 1 << (height - 1);
int left = 0, right = last_layer_count;
while (left < right) {
int middle = left + (right - left) / 2;
auto iter = root;
for (int i = 0, bit = 1 << (height - 2); i < height - 1; ++i) {
if ((bit & middle) == 0) {
iter = iter->left;
} else {
iter = iter->right;
}
bit >>= 1;
}
if (iter == nullptr) {
right = middle;
} else {
left = middle + 1;
}
}
return (1 << (height - 1)) - 1 + left;
}
};