0222. 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 2h nodes inclusive at the last level h.
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.
Follow up: Traversing the tree to count the number of nodes in the tree is an easy solution but with O(n) complexity. Could you find a faster algorithm?
Analysis¶
Complete binary tree has the special property: at the last level of the binary tree, the nodes will be populated from left to right. There could be empty/null from any points from the last level, but from the first empty/null node, all the rest to the right has to be empty/null as well.
If we are given a perfect binary tree, we can use 2^{height} - 1 to calculate the total number of nodes from the tree. The problem now breaks down to find starting at which position, the rest nodes of the leaf nodes become empty/null. We can use binary search to find the pivot node.
- start from left and right, and traverse \log(n) levels to see the heights of left and right.
- if left and right heights are the same, it means we already in perfect binary tree -> we can use the formula to calculate the number of nodes in the current subtree.
-
if left = right + 1, then the pivot should be somewhere between left and right, we should recursively call both.
-
Time: \log^2 (n)
- Space: \log(n) -> it takes \log(n) times to find the pivot thus it will call the recursion \log(n) times.
Code¶
/**
* 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) {
if (!root) return 0;
int l = 0, r = 0;
TreeNode *a = root, *b = root;
while (a) {
a = a -> left;
l++;
}
while (b) {
b = b -> right;
r++;
}
if (l == r) return pow(2, l) - 1;
return countNodes(root -> left) + countNodes(root -> right) + 1;
}
};