Merge branch 'master' of github.com:youngyangyang04/leetcode-master

problems/0222.完全二叉树的节点个数.md

@@ -447,7 +447,80 @@ 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;
+}
+```
+
+满二叉树
+```c
+int countNodes(struct TreeNode* root){
+    if(!root)
+        return 0;
+    int leftHeight = 0;
+    int rightHeight = 0;
+    struct TreeNode* rightNode = root->right;
+    struct TreeNode* leftNode = root->left;
+    //求出左子树深度
+    while(leftNode) {
+        leftNode = leftNode->left;
+        leftHeight++;
+    }
+
+    //求出右子树深度
+    while(rightNode) {
+        rightNode = rightNode->right;
+        rightHeight++;
+    }
+    //若左右子树深度相同，为满二叉树。结点个数为2^height-1
+    if(rightHeight == leftHeight) {
+        return (2 << leftHeight) - 1;
+    }
+    //否则返回左右子树的结点个数+1
+    return countNodes(root->right) + countNodes(root->left) + 1;
+}
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>