Update 0222.完全二叉树的节点个数.md

problems/0222.完全二叉树的节点个数.md

@@ -240,6 +240,71 @@ class Solution {
 
 Python：
 
+> 递归法：
+```python
+class Solution:
+    def countNodes(self, root: TreeNode) -> int:
+        return self.getNodesNum(root)
+        
+    def getNodesNum(self, cur):
+        if not cur:
+            return 0
+        leftNum = self.getNodesNum(cur.left) #左
+        rightNum = self.getNodesNum(cur.right) #右
+        treeNum = leftNum + rightNum + 1 #中
+        return treeNum
+```
+
+> 递归法：精简版
+```python
+class Solution:
+    def countNodes(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        return 1 + self.countNodes(root.left) + self.countNodes(root.right)
+```
+
+> 迭代法：
+```python
+import collections
+class Solution:
+    def countNodes(self, root: TreeNode) -> int:
+        queue = collections.deque()
+        if root:
+            queue.append(root)
+        result = 0
+        while queue:
+            size = len(queue)
+            for i in range(size):
+                node = queue.popleft()
+                result += 1 #记录节点数量
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+        return result
+```
+
+> 完全二叉树
+```python
+class Solution:
+    def countNodes(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        left = root.left
+        right = root.right
+        leftHeight = 0 #这里初始为0是有目的的，为了下面求指数方便
+        rightHeight = 0
+        while left: #求左子树深度
+            left = left.left
+            leftHeight += 1
+        while right: #求右子树深度
+            right = right.right
+            rightHeight += 1
+        if leftHeight == rightHeight:
+            return (2 << leftHeight) - 1 #注意(2<<1) 相当于2^2，所以leftHeight初始为0
+        return self.countNodes(root.left) + self.countNodes(root.right) + 1
+```
 
 Go：