标签:层遍历 next queue 入队 back pop solution time run
一、题目说明
题目102. Binary Tree Level Order Traversal,给一个二叉树,返回按层遍历的各节点,每层返回一个数组。
二、我的解答
这个题目是普通的二叉树按层遍历,没什么难度。代码如下:
class Solution{
    public:
        vector<vector<int>> levelOrder(TreeNode* root){
            vector<vector<int>> r;
            TreeNode* p;
            if(root ==NULL) return r;
            queue<TreeNode*> q;
            q.push(root);
            int curLevelNum = 1,nextLevelNum=0;
            while(! q.empty()){
                //出队列 
                vector<int> cur;
                for(int i=0;i<curLevelNum;i++){
                    p = q.front();
                    cur.push_back(p->val);
                    q.pop();
                    
                    //入队列
                    if(p->left !=NULL){
                        q.push(p->left);
                        nextLevelNum++;
                    } 
                    if(p->right !=NULL){
                        q.push(p->right);
                        nextLevelNum++;
                    }
                }
                curLevelNum = nextLevelNum;
                nextLevelNum = 0;
                r.push_back(cur);   
            }
            return r;
        }
};性能如下:
Runtime: 8 ms, faster than 60.42% of C++ online submissions for Binary Tree Level Order Traversal.
Memory Usage: 13.8 MB, less than 90.14% of C++ online submissions for Binary Tree Level Order Traversal.三、优化措施
不再做任何优化了。
刷题102. Binary Tree Level Order Traversal
标签:层遍历 next queue 入队 back pop solution time run
原文地址:https://www.cnblogs.com/siweihz/p/12266335.html