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