给定一个二叉树,返回其按层次遍历的节点值。(即逐层地,从左到右访问所有节点)
示例 1:

输入:root = [3,9,20,null,null,15,7]
输出:[[3],[9,20],[15,7]]
示例 2:
输入:root = [1]
输出:[[1]]
示例 3:
输入:root = []
输出:[]
提示:
- 树中节点数目在范围
[0, 2000]内 -1000 <= Node.val <= 1000
思路:
基本的树操作,可以用队列暂存一下节点访问
AC代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> result;
queue<TreeNode*> nodeQueue;
vector<int> level;
int size,i;
TreeNode* current;
if(root == NULL) return result;
nodeQueue.push(root);
while(!nodeQueue.empty()) {
level.clear();
size = nodeQueue.size();
for(i=0; i < size; i++) {
current = nodeQueue.front();
nodeQueue.pop();
level.push_back(current->val);
if(current->left) {
nodeQueue.push(current->left);
}
if(current->right) {
nodeQueue.push(current->right);
}
}
result.push_back(level);
}
return result;
}
};Code language: C++ (cpp)
稍快一些的做法是:
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
if (!root) return res;
queue<TreeNode*> nodeQueue{{root}};
while (!nodeQueue.empty()) {
res.emplace_back(); // 直接添加空vector,原地构造
for (int i = nodeQueue.size(); i > 0; --i) {
auto node = nodeQueue.front(); nodeQueue.pop();
res.back().push_back(node->val);
if (node->left) nodeQueue.push(node->left);
if (node->right) nodeQueue.push(node->right);
}
}
return res;
}
};Code language: C++ (cpp)