Related Topics: “树”: https://leetcode.com/tag/tree/ “广度优先搜索”: https://leetcode.com/tag/breadth-first-search/ “二叉树”: https://leetcode.com/tag/binary-tree/ Similar Questions: “二叉树的锯齿形层序遍历”: https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ “二叉树的层序遍历 II”: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ “二叉树的最小深度”: https://leetcode.com/problems/minimum-depth-of-binary-tree/ “二叉树的垂直遍历”: https://leetcode.com/problems/binary-tree-vertical-order-traversal/ “二叉树的层平均值”: https://leetcode.com/problems/average-of-levels-in-binary-tree/ “N 叉树的层序遍历”: https://leetcode.com/problems/n-ary-tree-level-order-traversal/ “二叉树的堂兄弟节点”: https://leetcode.com/problems/cousins-in-binary-tree/
Problem: Link to heading
给你一个二叉树，请你返回其按 层序遍历 得到的节点值。 （即逐层地，从左到右访问所有节点）。
示例：
二叉树：[3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回其层序遍历结果：
[
[3],
[9,20],
[15,7]
]
Solution: Link to heading
https://programmercarl.com/0102.二叉树的层序遍历.html
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
queue<TreeNode*> que;
if (root != NULL) que.push(root);
vector<vector<int>> result;
while (!que.empty()) {
int size = que.size();
vector<int> vec;
// 这里一定要使用固定大小size，不要使用que.size()，因为que.size是不断变化的
for (int i = 0; i < size; i++) {
TreeNode* node = que.front();
que.pop();
vec.push_back(node->val);
if (node->left) que.push(node->left);
if (node->right) que.push(node->right);
}
result.push_back(vec);
}
return result;
}
};