Create 0637-average-of-levels-in-binary-tree.cpp

Author: Abe0770

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

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+/*
+  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;
+    }
+};