{"id":142,"date":"2017-09-07T02:18:40","date_gmt":"2017-09-07T09:18:40","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=142"},"modified":"2018-07-10T18:55:03","modified_gmt":"2018-07-11T01:55:03","slug":"leetcode-637-average-of-levels-in-binary-tree","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/tree\/leetcode-637-average-of-levels-in-binary-tree\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 637. Average of Levels in Binary Tree"},"content":{"rendered":"<p><iframe loading=\"lazy\" title=\"LeetCode 637. Average of Levels in Binary Tree - \u82b1\u82b1\u9171 \u5237\u9898\u627e\u5de5\u4f5c EP14\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/3VljCEnwcdU?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share\" referrerpolicy=\"strict-origin-when-cross-origin\" allowfullscreen><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p><strong>Problem:<\/strong><\/p>\n<p>Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.<\/p>\n<p><b>Example 1:<\/b><\/p>\n<pre class=\"\"><b>Input:<\/b>\r\n    3\r\n   \/ \\\r\n  9  20\r\n    \/  \\\r\n   15   7\r\n<b>Output:<\/b> [3, 14.5, 11]\r\n<b>Explanation:<\/b>\r\nThe average value of nodes on level 0 is 3,  on level 1 is 14.5, and on level 2 is 11. \r\nHence return [3, 14.5, 11].<\/pre>\n<p>&nbsp;<\/p>\n<p><strong>Time Complexity:<\/strong><\/p>\n<p>O(n)<\/p>\n<p><strong>Space Complexity:<\/strong><\/p>\n<p>O(h)<\/p>\n<p><strong>Solution 1:<\/strong><\/p>\n<p>BFS<\/p>\n<pre class=\"lang:default decode:true\">\/\/ Author: Huahua\r\nclass Solution {\r\npublic:\r\n    vector&lt;double&gt; averageOfLevels(TreeNode* root) {\r\n        if(root == nullptr) return {};\r\n        vector&lt;double&gt; ans;\r\n        vector&lt;TreeNode*&gt; curr, next;\r\n        curr.push_back(root);\r\n        \r\n        \/\/ process every level one by one\r\n        while(!curr.empty()) {\r\n            long long sum = 0;\r\n            for(const auto&amp; node : curr) {\r\n                sum += node-&gt;val;\r\n                if (node-&gt;left) next.push_back(node-&gt;left);\r\n                if (node-&gt;right) next.push_back(node-&gt;right);\r\n            }\r\n            \r\n            ans.push_back(static_cast&lt;double&gt;(sum) \/ curr.size());\r\n            \r\n            curr.swap(next);\r\n            next.clear();\r\n        }\r\n        \r\n        return ans;\r\n    }\r\n};<\/pre>\n<p><strong>Solution 2:<\/strong><\/p>\n<p>DFS<\/p>\n<pre class=\"lang:default decode:true  \">\/\/ Author: Huahua\r\nclass Solution {\r\npublic:\r\n    vector&lt;double&gt; averageOfLevels(TreeNode* root) {\r\n        if(root == nullptr) return {};\r\n        vector&lt;pair&lt;long long, int&gt;&gt; sum_count;\r\n        vector&lt;double&gt; ans;\r\n        preorder(root, 0, sum_count);\r\n        for(const auto&amp; p : sum_count)\r\n            ans.push_back(static_cast&lt;double&gt;(p.first) \/ p.second);\r\n        return ans;\r\n    }\r\n    \r\nprivate:\r\n    void preorder(TreeNode* root, int depth, vector&lt;pair&lt;long long, int&gt;&gt;&amp; sum_count) {\r\n        if(root == nullptr) return;\r\n        if(depth&gt;=sum_count.size()) sum_count.push_back({0,0});\r\n        sum_count[depth].first += root-&gt;val;\r\n        ++sum_count[depth].second;\r\n        preorder(root-&gt;left, depth+1, sum_count);\r\n        preorder(root-&gt;right, depth+1, sum_count);\r\n    }\r\n};<\/pre>\n<p>&nbsp;<\/p>\n<p><strong>Related Problems:<\/strong><\/p>\n<ul>\n<li><a href=\"http:\/\/zxi.mytechroad.com\/blog\/leetcode\/leetcode-102-binary-tree-level-order-traversal\/\">[\u89e3\u9898\u62a5\u544a] LeetCode 102. Binary Tree Level Order Traversal<\/a><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>&nbsp; Problem: Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array. Example 1:&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[45],"tags":[34,30,33],"class_list":["post-142","post","type-post","status-publish","format-standard","hentry","category-tree","tag-bfs","tag-binary-tree","tag-dfs","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/142","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=142"}],"version-history":[{"count":4,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/142\/revisions"}],"predecessor-version":[{"id":3079,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/142\/revisions\/3079"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=142"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=142"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=142"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}