LeetCode 102. Binary Tree Level Order Traversal 

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<vector<int>> levelOrder(TreeNode* root) {
13         vector<vector<int>> res;
14         vector<int> level;
15         if(!root) return res;
16         
17         queue<TreeNode*> q1,q2;
18         q1.push(root);
19         while(!q1.empty()){
20             while(!q1.empty()){
21                 TreeNode* cur = q1.front();
22                 q1.pop();
23                 level.push_back(cur->val);
24                 if(cur->left) q2.push(cur->left);
25                 if(cur->right) q2.push(cur->right);
26             }
27             res.push_back(level);
28             level.clear();
29             q1.swap(q2);
30             
31         }
32         return res;
33     }
34 };

和之前的某题类似，维护两个queue（不是stack因为不是zigzag），q1存当前层的节点，q2存下一层节点。

 

另一种写法：（java）

 1 public class Solution {
 2     public List<List<Integer>> levelOrder(TreeNode root) {
 3         List<List<Integer>> res = new ArrayList<List<Integer>>();
 4         if (root == null)
 5             return res;
 6         Queue<TreeNode> q = new LinkedList<TreeNode>();
 7         q.offer(root);
 8         while (!q.isEmpty()) {
 9             List<Integer> tmp = new ArrayList<Integer>();
10             int size = q.size();
11             for (int i = 0; i < size; ++i) {
12                 TreeNode t = q.poll();
13                 if (t.left != null)
14                     q.offer(t.left);
15                 if (t.right != null)
16                     q.offer(t.right);
17                 tmp.add(t.val);
18             }
19             res.add(tmp);
20         }
21         return res;
22     }
23 }

 

新写法：（留坑）

 1 public class Solution {
 2     public List<List<Integer>> levelOrder(TreeNode root) {
 3         List<List<Integer>> res = new ArrayList<List<Integer>>();
 4         dfs(res,root,0);
 5         return res;
 6     }
 7     public void dfs(List<List<Integer>> list,TreeNode node,int deep){
 8         if(node==null)return;
 9         if(list.size()==deep)
10             list.add(new ArrayList<Integer>());
11         list.get(deep).add(node.val);
12         dfs(list, node.left, deep+1);
13         dfs(list, node.right, deep+1);    
14     }
15 }

 

posted @ 2016-03-02 20:28  co0oder  阅读(151)  评论(0)  编辑  收藏  举报