Preface
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Problem
Given the root node root of a binary tree, return its level order traversal (i.e., visit all nodes level by level, from left to right).
Example 1
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输入：root = [3,9,20,null,null,15,7]
输出：[[3],[9,20],[15,7]]
Example 2
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输入：root = [1]
输出：[[1]]
Example 3
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输入：root = []
输出：[]
Implementation Code
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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>> ret;
if(root == nullptr) { return ret; }
queue <TreeNode *> q;
q.push(root);
while(!q.empty()) {
int levelSize = q.size();
ret.push_back(vector<int>());
for (int i = 1; i <= levelSize; ++i) {
auto node = q.front();
q.pop();
ret.back().push_back(node->val);
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
}
return ret;
}
};