Description
Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: [[3],[9,20],[15,7]]
Example 2:
Input: root = [1] Output: 1
Example 3:
Input: root = [] Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -1000 <= Node.val <= 1000
Code
Run BFS, collect node along the way.
/**
* 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) {
if(!root) return {};
queue<TreeNode*> q;
q.push(root);
vector<vector<int>> res;
vector<int> level = {};
while(!q.empty()) {
int n = q.size();
level.clear();
for(int i = 0; i < n; i++) {
auto node = q.front();
level.push_back(node->val);
q.pop();
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
res.push_back(level);
}
return res;
}
};