目录

一、题目

二、代码

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一、题目

102. 二叉树的层序遍历 - 力扣（LeetCode）

 

二、代码

• 主要应用到了两个队列，一个队列存放数据，一个队列存放对应所在的第几层

/*** 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:int Number_of_layers(TreeNode* root)//计算二叉树的层数{if(root == nullptr){return 0;}int left = Number_of_layers(root->left);int right = Number_of_layers(root->right);return left>right?left+1:right+1;}vector<vector<int>> levelOrder(TreeNode* root) {queue<TreeNode*>d1;//存放结点queue<int>d2;//存放对应在第几层int Number_layers = Number_of_layers(root);vector<vector<int>>data(Number_layers);//提前指定二维vector的行数if(root == nullptr)return data;d1.push(root);d2.push(0);while(!d1.empty())//层序遍历{TreeNode*front = d1.front();int i = d2.front();data[i].push_back(front->val);d1.pop();d2.pop();if(front->left!=nullptr){d1.push(front->left);d2.push(i+1);}if(front->right!=nullptr){d1.push(front->right);d2.push(i+1);}}return data;}
};