按照广度优先的思想,设置两个队列,一个队列存放当前level的所有节点,另外一个队列存放当前节点的子节点。
/**
* @author johnsondu
* @problem Binary Tree Level Order Traversal
* @url https://leetcode.com/problems/binary-tree-level-order-traversal/
* @timeComlexity O(n)
* @spaceComplexity O(n)
* @strategy bfs
*/
/**
* 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>> ans;
queue<TreeNode*> q1, q2;
if(root == NULL) return ans;
q1.push(root);
vector<int> layer;
while(!q1.empty()) {
layer.clear();
while(!q1.empty()) {
TreeNode* cur = q1.front();
q1.pop();
layer.push_back(cur->val);
if(cur->left) {
q2.push(cur->left);
}
if(cur->right) {
q2.push(cur->right);
}
}
ans.push_back(layer);
q1 = q2;
while(!q2.empty()) q2.pop();
}
return ans;
}
};