637. Average of Levels in Binary Tree

最新推荐文章于 2020-02-11 14:21:39 发布

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本文介绍了一种算法，该算法接收一棵非空二叉树作为输入，并返回一个数组，数组中的每个元素代表对应层级节点值的平均数。通过广度优先搜索（BFS）遍历二叉树，统计每一层的节点数量及节点值总和，进而计算平均值。

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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.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<double> res;
    void bfs(TreeNode* root){
        deque<TreeNode*> d(NULL);
        d.push_back(root);
        double tmp = 0;
        int fatherNum = 1, childNum = 0, cnt = 0;
        while(1){            
            if(root->left != NULL){
                d.push_back(root->left);
                childNum++;
            }
            if(root->right != NULL){
                d.push_back(root->right);
                childNum++;
            }              
            tmp += d[0]->val;
            d.pop_front();                        
            cnt++;
            if(cnt == fatherNum){
                res.push_back(tmp / cnt);
                tmp = 0;
                cnt = 0;
                fatherNum = childNum;
                childNum = 0;
            }            
            if(d.empty())
                break;
            root = d[0];
        }
    }
    vector<double> averageOfLevels(TreeNode* root) {
        if(root == NULL) return NULL;         
        bfs(root);
        return res;        
    }
};