核心思想:通过一个队列保存每一层的所有结点,之后逐层遍历,并且将每个结点的左右儿子结点存入队中。
/**
* 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) return res;//如果树为空直接返回
queue<TreeNode*> q;
q.push(root);
TreeNode* p; // 追踪当前的结点
while(!q.empty()){
int levelSize = q.size(); //统计当前层的结点数量
vector<int> currentLevel; //开一个vector存当前层的所有结点
for(int i = 0; i < levelSize; i ++){
p = q.front();
q.pop();
currentLevel.push_back(p->val);
if(p->left) q.push(p->left);
if(p->right) q.push(p->right);
}
//将当前层的vector存入答案的vector res中
res.push_back(currentLevel);
}
return res;
}
};