leetcode 102 二叉树的层次遍历 (Binary Tree Level Order Traversal)

我的方法。每个队列保存一层的node：

/**
 * 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*> qu;
        if(!root){
            return res;
        }
        qu.push(root);
        soln(res,qu);
        return res;
    }
    
    void soln(vector<vector<int>>& res, queue<TreeNode*>& qu1){
        if(qu1.empty()){
            return;
        }
        queue<TreeNode*> qu2;
        vector<int> tem;
        while(!qu1.empty()){
            
            tem.push_back(qu1.front()->val);
            if(qu1.front()->left){
                qu2.push(qu1.front()->left);
            }
            if(qu1.front()->right){
                qu2.push(qu1.front()->right);
            }
            qu1.pop();
            
        }
        res.push_back(tem);
        soln(res,qu2);
        
    }
    
};

8ms 。

可以用递归。

为什么非得一层加完了再加下一层呢？没有必要。

class Solution {  
    private:  
    vector<vector<int>>res;  
public:  
  void levelOrder(struct TreeNode* root,int level){  
     if(root==NULL)return ;  
     if(level==res.size()){  
     vector<int> v;  
     res.push_back(v);  
     }  
     res[level].push_back(root->val);  
     levelOrder(root->left,level+1);  
     levelOrder(root->right,level+1);  
  }  
    vector<vector<int>> levelOrder(TreeNode* root) {  
      levelOrder(root,0);  
        return res;    
    }  
};  

4ms。 

只用一个队列，来自leetcode。

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if (root == nullptr) return vector<vector<int>>{};
        vector<vector<int>> res;
        vector<int> row;
        TreeNode *last, *pre_last = root, *cur = root;
        queue<TreeNode*> nodes;
        nodes.push(cur);
        while (!nodes.empty()) {
            cur = nodes.front();
            row.push_back(cur->val);
            if (cur->left) {
                nodes.push(cur->left);
                last = cur->left;
            }
            if (cur->right) {
                nodes.push(cur->right);
                last = cur->right;
            }
            if (cur == pre_last) {
                res.push_back(row);
                row.clear();
                pre_last = last;
            }
            nodes.pop();
        }
        return res;
    }
};

4ms。