Given a complete binary tree, count the number of nodes.
Note:
Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, 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:
Input: 1 / \ 2 3 / \ / 4 5 6 Output: 6
思路:
我们使用分层遍历的思想完成该题,虽然理论上可以不完全遍历所有的节点,但是实际上O(n)算法已经算效率非常高的一种算法了。代码如下:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int countNodes(TreeNode* root) {
if(root == nullptr) return 0;
list<TreeNode *> s;
s.push_back(root);
int count(0);
while (!s.empty()) {
TreeNode *top = s.front();
s.pop_front();
if (top->left) s.push_back(top->left);
if (top->right) s.push_back(top->right);
count++;
}
return count;
}
};