From 61044a1f94583e817dbf799e8d5c9b59da800e8b Mon Sep 17 00:00:00 2001
From: Arthur <gpan20020205@gmail.com>
Date: Mon, 8 Nov 2021 12:20:52 +0000
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---
 problems/0222.完全二叉树的节点个数.md | 44 +++++++++++++++++++++++++++
 1 file changed, 44 insertions(+)

diff --git a/problems/0222.完全二叉树的节点个数.md b/problems/0222.完全二叉树的节点个数.md
index dc09985d..0826110f 100644
--- a/problems/0222.完全二叉树的节点个数.md
+++ b/problems/0222.完全二叉树的节点个数.md
@@ -449,7 +449,51 @@ var countNodes = function(root) {
 };
 ```
 
+## C:
+递归法
+```c
+int countNodes(struct TreeNode* root) {
+    //若传入结点不存在，返回0
+    if(!root)
+        return 0;
+    //算出左右子树的结点总数
+    int leftCount = countNodes(root->left);
+    int rightCount = countNodes(root->right);
+    //返回左右子树结点总数+1
+    return leftCount + rightCount + 1;
+}
 
+int countNodes(struct TreeNode* root){
+    return getNodes(root);
+}
+```
+
+迭代法
+```c
+int countNodes(struct TreeNode* root){
+    //记录结点总数
+    int totalNum = 0;
+    //开辟栈空间
+    struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 100);
+    int stackTop = 0;
+    //如果root结点不为NULL，则将其入栈。若为NULL，则不会进入遍历，返回0
+    if(root)
+        stack[stackTop++] = root;
+    //若栈中有结点存在，则进行遍历
+    while(stackTop) {
+        //取出栈顶元素
+        struct TreeNode* tempNode = stack[--stackTop];
+        //结点总数+1
+        totalNum++;
+        //若栈顶结点有左右孩子，将它们入栈
+        if(tempNode->left)
+            stack[stackTop++] = tempNode->left;
+        if(tempNode->right)
+            stack[stackTop++] = tempNode->right;
+    }
+    return totalNum;
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)