Merge pull request #2770 from Abe0770/main

Author: felivalencia3

Diff for: cpp/0103-binary-tree-zigzag-level-order-traversal.cpp

@@ -0,0 +1,56 @@
+/*
+  Given the root of a binary tree, return the zigzag level order traversal of its nodes' values. 
+  (i.e., from left to right, then right to left for the next level and alternate between).
+
+  Ex. Input: root = [3,9,20,null,null,15,7]
+      Output: [[3],[20,9],[15,7]]
+
+  Time  : O(N)
+  Space : O(N)
+*/
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
+        vector <vector <int>> v;
+        if(root == NULL)
+            return v;
+      
+        queue <TreeNode *> q;
+        q.push(root);
+        bool changeDirection = true;
+      
+        while(!q.empty()) {
+            vector <int> v1;
+            int siz = q.size();
+          
+            for(int i = 0 ; i < siz ; i++) {
+                if(changeDirection)
+                    v1.push_back(q.front() -> val);
+                else 
+                    v1.insert(v1.begin(), q.front() -> val);
+              
+                if(q.front() -> left != NULL)
+                    q.push(q.front() -> left);
+                if(q.front() -> right != NULL)
+                    q.push(q.front() -> right);
+              
+                q.pop();
+            }
+            changeDirection = !changeDirection;
+            v.push_back(v1);
+        }
+        return v;
+    }
+};

Diff for: cpp/0116-populating-next-right-pointers-in-each-node.cpp

@@ -0,0 +1,69 @@
+/*
+  You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:
+  
+  struct Node {
+    int val;
+    Node *left;
+    Node *right;
+    Node *next;
+  }
+  
+  Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
+  Initially, all next pointers are set to NULL.
+
+  Ex. Input: root = [1,2,3,4,5,6,7]
+      Output: [1,#,2,3,#,4,5,6,7,#]
+      Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node,
+      just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
+
+  Time  : O(N)
+  Space : O(N)
+*/
+
+/*
+// Definition for a Node.
+class Node {
+public:
+    int val;
+    Node* left;
+    Node* right;
+    Node* next;
+
+    Node() : val(0), left(NULL), right(NULL), next(NULL) {}
+
+    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
+
+    Node(int _val, Node* _left, Node* _right, Node* _next)
+        : val(_val), left(_left), right(_right), next(_next) {}
+};
+*/
+
+class Solution {
+public:
+    Node* connect(Node* root) {
+        if(root == NULL || (root -> right == NULL && root -> left == NULL))
+            return root;
+        queue <Node*>q;
+        q.push(root);
+
+        while(!q.empty()) {
+            int size = q.size();
+            Node* temp = NULL;
+
+            for(int i = 0 ; i < size ; i++) {
+                Node* front = q.front();
+
+                if(temp != NULL)
+                    temp -> next = front;
+                temp = front;
+
+                if(front -> left) {
+                    q.push(front -> left);
+                    q.push(front -> right);
+                }
+                q.pop();
+            }
+        }
+        return root;
+    }
+};

Diff for: cpp/0117-populating-next-right-pointers-in-each-node-ii.cpp

@@ -0,0 +1,68 @@
+/*
+  Given a binary tree
+
+  struct Node {
+    int val;
+    Node *left;
+    Node *right;
+    Node *next;
+  }
+  
+  Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
+  
+  Initially, all next pointers are set to NULL.
+
+  Ex. Input: root = [1,2,3,4,5,null,7]
+      Output: [1,#,2,3,#,4,5,7,#]
+      Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node,
+      just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.
+
+  Time  : O(N)
+  Space : O(1)
+*/
+
+/*
+// Definition for a Node.
+class Node {
+public:
+    int val;
+    Node* left;
+    Node* right;
+    Node* next;
+
+    Node() : val(0), left(NULL), right(NULL), next(NULL) {}
+
+    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}
+
+    Node(int _val, Node* _left, Node* _right, Node* _next)
+        : val(_val), left(_left), right(_right), next(_next) {}
+};
+*/
+
+class Solution {
+public:
+    Node* connect(Node* root) {
+        if(root == NULL)
+            return NULL;
+        Node* parent = root;
+
+        while(parent) {
+            Node* newNode = new Node;
+            Node* temp = newNode;
+
+            while(parent) {
+                if(parent -> left) {
+                    temp -> next = parent -> left;
+                    temp = temp -> next;
+                }
+                if(parent -> right) {
+                    temp -> next = parent -> right;
+                    temp = temp -> next;
+                }
+                parent = parent -> next;
+            }
+            parent = newNode -> next;
+        }
+        return root;
+    }
+};

Diff for: cpp/0637-average-of-levels-in-binary-tree.cpp

@@ -0,0 +1,50 @@
+/*
+  Given the root of a binary tree, return the average value of the nodes on each level in the form of an array. 
+  Answers within 10-5 of the actual answer will be accepted. 
+
+  Ex. Input: root = [3,9,20,null,null,15,7]
+      Output: [3.00000,14.50000,11.00000]
+      Explanation: The average value of nodes on level 0 is 3, on level 1 is 14.5, and on level 2 is 11.
+      Hence return [3, 14.5, 11].
+
+  Time  : O(N)
+  Space : O(N)
+*/
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    vector<double> averageOfLevels(TreeNode* root) {
+        vector <double> v;
+        if(root == NULL) 
+            return v;
+        queue <TreeNode *> q;
+        q.push(root);
+
+        while(!q.empty()) {
+            double sum = 0, count = 0;
+            int siz = q.size();
+            for(int i = 0 ; i < siz ; i++) {
+                sum += q.front() -> val;
+                if(q.front() -> left)
+                    q.push(q.front() -> left);
+                if(q.front() -> right) 
+                    q.push(q.front() -> right);
+                ++count;
+                q.pop();
+            }
+            v.push_back(sum / count);
+        }
+        return v;
+    }
+};