二叉树的层序遍历(102)
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