題目
https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
思路
隊列實現BFS;使用隊列來存放每層的節點,從第一層開始檢索,如果有孩子就壓入。隊列:先進先出,遍歷隊列的front,即可實現BFS
AC代碼
/**
* 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>> res;
queue<TreeNode *>q;
if(root == NULL) return res;
q.push(root);
while (!q.empty()){
vector<int>temp;
int len = q.size();
for(int i = 0;i < len;i++){
//注意這個len不能直接寫成q.size(),因爲後面有新的插入,會重新計算q.size();
TreeNode * node = q.front();
q.pop();
temp.push_back(node->val);
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
res.push_back(temp);
}
return res;
}
};