2021.2.3 刷题(层序遍历2）

题目链接：https://leetcode-cn.com/problems/binary-tree-right-side-view/
题目描述：
1.二叉树的右视图

思路：二叉树层序遍历的时候，判断是否遍历到单层的最后面的元素，如果是，就放进result数组中。

/**
 * 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<int> rightSideView(TreeNode* root) {
        vector<int> result;
        queue<TreeNode *> que;
        if(root != nullptr)
            que.push(root);
        while(!que.empty())
        {
            int currentLevel = que.size();
            TreeNode *node;
            for(int i = 0; i < currentLevel; i++)
            {
                node = que.front();
                que.pop();
                if(node->left) que.push(node->left);
                if(node->right) que.push(node->right);
            }
            result.push_back(node->val);  //本层的最后一个节点
        }
        return result;
    }
};

2.二叉树的层平均值
题目链接：https://leetcode-cn.com/problems/average-of-levels-in-binary-tree/
题目描述：给定一个非空二叉树, 返回一个由每层节点平均值组成的数组。

题解：

/**
 * 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> averageOfLevels(TreeNode* root) {
        vector<double> result;
        queue<TreeNode *> que;
        if(root != NULL)
            que.push(root);
        while(!que.empty())
        {
            int currentLevel = que.size();
            double sum = 0;
            for(int i = 0; i < currentLevel; i++)
            {
                TreeNode *node = que.front();
                que.pop();
                sum += node->val;
                if(node->left) que.push(node->left);
                if(node->right) que.push(node->right);

            }
            result.push_back(sum / currentLevel);
        }
        return result;
    }
};

3.N叉树的层序遍历
题目链接：https://leetcode-cn.com/problems/n-ary-tree-level-order-traversal/
题目描述：给定一个 N 叉树，返回其节点值的层序遍历。（即从左到右，逐层遍历）。


解题：

/*
// Definition for a Node.
class Node {
public:
    int val;
    vector<Node*> children;

    Node() {}

    Node(int _val) {
        val = _val;
    }

    Node(int _val, vector<Node*> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> result;
        queue<Node *> que;
        if(root != NULL)
            que.push(root);
        while(!que.empty())
        {
            vector<int> vec;
            int currentLevel = que.size();
            for(int i = 0; i < currentLevel; i++)
            {
                Node *p = que.front();
                que.pop();
                vec.push_back(p->val);
                for(int i = 0; i < p->children.size(); i++)  //节点的孩子入队
                {
                    if(p->children[i])
                        que.push(p->children[i]);
                }
            }
            result.push_back(vec);
        }
        return result;
    }
};