sol 114, 116, 222 and 243

Author: ioMayday

c_code/114_flatten_1.c

@@ -0,0 +1,37 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     struct TreeNode *left;
+ *     struct TreeNode *right;
+ * };
+ */
+
+
+// 递归写法
+void flatten(struct TreeNode* root)
+{
+    if (root == NULL) return;
+
+    // 先得到两个flatten后的左右子树
+    flatten(root->left);
+    flatten(root->right);
+
+    // 重新接上左右子树
+    struct TreeNode* left = root->left;
+    struct TreeNode* right = root->right;
+
+    // 接上左子树
+    root->left = NULL;
+    root->right = left;
+
+    // 找到原左子树flatteng的末尾节点
+    struct TreeNode* p = root;  //直接从root开始赋值,比root->right赋值要好,省去一个NULL判断
+    while (p->right != NULL) { 
+        p = p->right;
+    }
+    // 接上原右子树
+    p->right = right; 
+
+    return;
+}

c_code/114_flatten_2.c

@@ -0,0 +1,105 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     struct TreeNode *left;
+ *     struct TreeNode *right;
+ * };
+ */
+
+// 自己的方法,利用先序方法重新建立个树,调试通过
+
+struct TreeNode {
+    int val;
+    struct TreeNode *left;
+    struct TreeNode *right;
+};
+
+void firstTraverse1(struct TreeNode* root)
+{
+    if (root == NULL) return;
+
+    printf("%d ", root->val);
+    firstTraverse1(root->left);
+    firstTraverse1(root->right);
+    return;
+}
+
+void firstTraverse2(struct TreeNode* root)
+{
+    if (root == NULL) return;
+
+    printf("%d ", root->val);
+    firstTraverse2(root->right);
+    return;
+}
+
+
+struct TreeNode* g_pre;
+void firstFlatten(struct TreeNode* root)
+{
+    if (root == NULL) return;
+    
+    // 先生成一个节点
+    struct TreeNode* node = (struct TreeNode *)malloc(sizeof(struct TreeNode));
+    node->val = root->val;  // copy root
+    node->left = NULL;
+    g_pre->right = node;
+    g_pre = node;
+
+    firstFlatten(root->left);
+    firstFlatten(root->right);
+
+    return;
+}
+
+// 自己的方法,利用先序遍历来连接节点
+void flatten(struct TreeNode* root)
+{
+    struct TreeNode* newRoot = (struct TreeNode *)malloc(sizeof(struct TreeNode));
+    newRoot->val = -1;
+    newRoot->left = NULL;
+    newRoot->right = NULL;
+    g_pre = newRoot;
+
+    firstTraverse1(root);  // left and right
+    printf("\n");
+
+    firstFlatten(root);
+    root = newRoot->right;  // 缺点是,由于接口限制,单独创立的next副本无法返回;
+    g_pre->right = NULL;    // 处理完后,所有节点后,最末尾的节点设为NULL
+    
+    firstTraverse2(root);  // only right
+    
+    return;
+}
+
+
+int main(void)
+{
+    struct TreeNode n1, n2, n3, n4, n5, n6;
+    n1.val = 1;
+    n2.val = 5;
+    n3.val = 6;
+    n4.val = 7;
+    n5.val = 8;
+    n6.val = 9;
+    n1.left = &n2;
+    n1.right = &n5;
+    n2.left = &n3;
+    n2.right = &n4;
+    n3.left = NULL;
+    n3.right = NULL;
+    n4.left = NULL;
+    n4.right = NULL;
+    n5.left = NULL;
+    n5.right = &n6;
+    n6.left = NULL;
+    n6.right = NULL;
+
+    flatten(&n1);
+
+    // while (1) { ; }
+
+    return 0;
+}

c_code/116_connect.c

@@ -0,0 +1,34 @@
+/**
+ * Definition for a Node.
+ * struct Node {
+ *     int val;
+ *     struct Node *left;
+ *     struct Node *right;
+ *     struct Node *next;
+ * };
+ */
+
+// 参考: 此题不熟，需要多练
+// https://labuladong.gitbook.io/algo/shu-ju-jie-gou-xi-lie/shou-ba-shou-shua-er-cha-shu-xun-lian-di-gui-si-wei/er-cha-shu-xi-lie-1
+
+void connectTwoNodes(struct Node* node1, struct Node* node2) 
+{
+    if (node1 == NULL || node2 == NULL) return;
+
+    // 连接左右两个节点
+    node1->next = node2;
+
+    // 连接同一父节点下两个节点
+    connectTwoNodes(node1->left, node1->right);
+    connectTwoNodes(node2->left, node2->right);
+    // 连接不同父节点下两个相邻节点
+    connectTwoNodes(node1->right, node2->left);
+}
+
+struct Node* connect(struct Node* root) {
+	if (root == NULL) return root;
+
+    connectTwoNodes(root->left, root->right);
+
+    return root;
+}

c_code/222_countNodes.c

@@ -0,0 +1,32 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     struct TreeNode *left;
+ *     struct TreeNode *right;
+ * };
+ */
+
+
+int countNodes(struct TreeNode* root){  // 可计算任意二叉树节点总数(包含完全二叉树（满二叉树）)
+    if (root == NULL) return 0;
+    
+    // 满二叉树
+    int hl = 1, hr = 1;
+    struct TreeNode* l = root->left;
+    struct TreeNode* r = root->right;
+    while (l != NULL) { // 右子树高度
+        l = l->left;
+        hl++;
+    }
+    while (r != NULL) { //左子树高度
+        r = r->right;
+        hr++;
+    }
+    if (hl == hr) {  // 若为满二叉树
+        return pow(2, hl) - 1; // 满二叉树节点计算：2^depth - 1
+    }
+
+    // 普通二叉树
+    return 1 + countNodes(root->left) + countNodes(root->right);
+}

c_code/243_isPalindrome.c renamed to c_code/234_isPalindrome.c

@@ -19,4 +19,13 @@
 **LeetCode：二叉树类递归题** <br>
 226.翻转二叉树（简单）  
 114.二叉树展开为链表（中等）  
-116.填充每个节点的下一个右侧节点指针（中等）  
+116.填充每个节点的下一个右侧节点指针（中等）  || 此题不熟，需要多练  
+226  
+654  
+105  
+106  
+222. 完全二叉树的节点个数  
+
+
+# 二分法思路训练题目搜集
+34. 在排序数组中查找元素的第一个和最后一个位置