LeetCode 102 Binary Tree Level Order Traversal 层序BFS
作者：互联网
Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).
Solution
用 \(BFS\) 即可，每次将该层的节点 \(pop\)， 然后 \(push\) 其子数的节点
点击查看代码
/**
* 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 {
private:
vector<vector<int>> ans;
queue<TreeNode*> q;
public:
vector<vector<int>> levelOrder(TreeNode* root) {
if(!root) return ans;
q.push(root);
while(!q.empty()){
int sz = q.size();
vector<int> tmp;
for(int i=0;i<sz;i++){
auto f = q.front(); q.pop();
tmp.push_back(f->val);
if(f->left)q.push(f->left);
if(f->right)q.push(f->right);
}
ans.push_back(tmp);
}
return ans;
}
};
标签：Binary,right,TreeNode,Level,int,层序,nullptr,push,left 来源： https://www.cnblogs.com/xinyu04/p/16513776.html