58.完全二叉树的节点个数

O(n)时间复杂度解法

class Solution {public int countNodes(TreeNode root) {int num=0;if(root==null){return num;}Queue<TreeNode> qu=new LinkedList<>();qu.offer(root);TreeNode tn;int le;while(!qu.isEmpty()){le=qu.size();while(le>0){tn=qu.poll();if(tn.left!=null){qu.offer(tn.left);}if(tn.right!=null){qu.offer(tn.right);}le--;num++;}}return num;}
}

class Solution(object):def countNodes(self, root):num=0if root is None:return numqu=collections.deque()qu.append(root)while qu:le=len(qu)while le>0:tn=qu.popleft()if tn.left:qu.append(tn.left)if tn.right:qu.append(tn.right)le-=1num+=1return num

以下解法来自代码随想录

class Solution {public int countNodes(TreeNode root) {if(root==null){return 0;}return countNodes(root.left)+countNodes(root.right)+1;}
}
//​解法来自代码随想录

class Solution(object):def countNodes(self, root):if root is None:return 0return self.countNodes(root.left)+self.countNodes(root.right)+1
#​以下解法来自代码随想录