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力扣102题:二叉树的层序遍历
2022-07-06 23:59:00 【瀛台夜雪】
力扣102题:二叉树的层序遍历
题目描述
给你二叉树的根节点 root
,返回其节点值的 层序遍历 。 (即逐层地,从左到右访问所有节点)。
输入输出样例
输入:root = [3,9,20,null,null,15,7]
输出:[[3],[9,20],[15,7]]
输入:root = [1]
输出:[[1]]
输入:root = []
输出:[]
解法1,使用队列利用迭代的方式
vector<vector<int>> levelOrder(TreeNode* root)
{
vector<vector<int>>res;
if(!root)
{
return res;
}
//建立队列
queue<TreeNode *>que;
que.push(root);
vector<int>tempList;
while(!que.empty())
{
int length=que.size();
res.push_back({
});
for(int i=0;i<length;i++)
{
TreeNode * temp=que.front();
que.pop();
// cout<<temp->val<<" ";
res.back().push_back(temp->val);
if(temp->left)
{
que.push(temp->left);
}
if(temp->right)
{
que.push(temp->right);
}
}
}
return res;
}
解法二,使用递归的方式进行
vector<vector<int>>nums;
//使用递归的方法进行解决
void dns(TreeNode *root,int lever)
{
if(!root)
{
return;
}
if(nums.size()==lever)
{
nums.push_back({
});
}
nums[lever].push_back(root->val);
dns(root->left,lever+1);
dns(root->right,lever+1);
}
vector<vector<int>> levelOrder2(TreeNode* root)
{
dns(root,0);
return nums;
}
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