102. 二叉树的层序遍历 - 力扣（LeetCode）

解法：

/*** 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<vector<int>> levelOrder(TreeNode* root) {vector<vector<int>> res;if (root == nullptr) {return res;}queue<TreeNode* >  q;q.push(root);while(!q.empty()) {uint32_t n = q.size();vector<int> v;v.reserve(n);for (int i = 0; i < n; i++) {TreeNode * t = q.front();q.pop();v.push_back(t->val);if (t->left != nullptr) {q.push(t->left);}if (t->right != nullptr) {q.push(t->right);}}res.push_back(std::move(v));}return res;}
};

总结：计算时间复杂度O(N)，空间复杂度O(N)。相关题目：二叉树前序遍历（144）、中序遍历（94）、后序遍历（145）-CSDN博客