leetcode102

本题是广度优先遍历（BFS）实现树的层次遍历，使用队列实现。

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        queue<TreeNode*>q;                
        if (root != NULL)
        {
            q.push(root);
            while (!q.empty())
            {
                vector<int> tmp;
                vector<TreeNode*> T;
                while (!q.empty())
                {
                    TreeNode* t = q.front();
                    q.pop();
                    tmp.push_back(t->val);
                    if (t->left != NULL)
                    {
                        T.push_back(t->left);
                    }
                    if (t->right != NULL)
                    {
                        T.push_back(t->right);
                    }
                }                
                res.push_back(tmp);
                for (auto x : T)
                {
                    q.push(x);
                }
            }
        }
        return res;
    }
};

补充一个python的实现：

 1 class Solution:        
 2     def lOrder(self,temp,result):
 3         count = len(temp)
 4         newary = []
 5         while count > 0:
 6             top = temp.pop(0)
 7             newary.append(top.val)
 8             count -= 1
 9             if top.left != None:
10                 temp.append(top.left)
11             if top.right != None:
12                 temp.append(top.right)
13         if len(newary) > 0:
14             result.append(newary)
15         if len(temp) > 0:
16             self.lOrder(temp,result)
17         
18     def levelOrder(self, root: TreeNode) -> List[List[int]]:
19         result = []
20         temp = []
21         if root != None:
22             temp.append(root)
23             self.lOrder(temp,result)
24         return result

posted on 2018-10-03 10:03 Sempron2800+ 阅读(160) 评论(0) 编辑收藏举报