Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.
Example 1:
Input: 3 / \ 9 20 / \ 15 7 Output: [3, 14.5, 11] 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].
Note:
- The range of node's value is in the range of 32-bit signed integer.
这道题让我们求一个二叉树每层的平均值,那么一看就是要进行层序遍历了,直接上queue啊,如果熟悉层序遍历的方法,那么这题就没有什么难度了,直接将每层的值累计加起来,除以该层的结点个数,存入结果res中即可,参见代码如下:
解法一:
public: vector<double> averageOfLevels(TreeNode* root) { if (!root) return {}; vector<double> res; queue<TreeNode*> q{{root}}; while (!q.empty()) { int n = q.size(); double sum = 0; for (int i = 0; i < n; ++i) { TreeNode *t = q.front(); q.pop(); sum += t->val; if (t->left) q.push(t->left); if (t->right) q.push(t->right); } res.push_back(sum / n); } return res; } };
下面这种方法虽然是利用的递归形式的先序遍历,但是其根据判断当前层数level跟结果res中已经初始化的层数之间的关系对比,能把当前结点值累计到正确的位置,而且该层的结点数也自增1,这样我们分别求了两个数组,一个数组保存了每行的所有结点值,另一个保存了每行结点的个数,这样对应位相除就是我们要求的结果了,参见代码如下:
解法二:
public: vector<double> averageOfLevels(TreeNode* root) { vector<double> res, cnt; helper(root, 0, cnt, res); for (int i = 0; i < res.size(); ++i) { res[i] /= cnt[i]; } return res; } void helper(TreeNode* node, int level, vector<double>& cnt, vector<double>& res) { if (!node) return; if (res.size() <= level) { res.push_back(0); cnt.push_back(0); } res[level] += node->val; ++cnt[level]; helper(node->left, level + 1, cnt, res); helper(node->right, level + 1, cnt, res); } };
参考资料:
https://discuss.leetcode.com/topic/95567/java-solution-using-dfs-with-full-comments
本文转自博客园Grandyang的博客,原文链接:[LeetCode] Average of Levels in Binary Tree 二叉树的层平均值
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时间: 2024-09-23 12:31:20