LeetCode OJ：Binary Tree Right Side View（右侧视角下的二叉树）

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example:
Given the following binary tree,

   1            <---

 /   \

2     3         <---

 \     \

  5     4       <---

You should return [1, 3, 4].

其实题目的意思就是相当于二叉树每一行的最右边的一个元素，那么简单，先bfs后取每一个数组的最后一位数就可以了，代码如下：

 /**

  * 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<int> rightSideView(TreeNode* root) {

         int dep = -;

         bfs(root, dep + );

         vector<int> res;

         for(int i = ; i < ret.size(); ++i){

             res.push_back(ret[i][ret[i].size() - ]);

         }

         return res;

     }

     void bfs(TreeNode * root, int depth)

     {

         if(root == NULL) return;

         if(depth < ret.size()){

             ret[depth].push_back(root->val);

         }else{

             vector<int>tmp;

             ret.push_back(tmp);

             ret[depth].push_back(root->val);

         }

         if(root->left)

             bfs(root->left, depth + );

         if(root->right)

             bfs(root->right, depth + );

     }

 private:

     vector<vector<int>> ret;

 };

java版本的代码如下所示，同样是BFS之后再取最后一位组成一个数组：

 public class Solution {

     public List<Integer> rightSideView(TreeNode root) {

         List<List<Integer>> ret = new ArrayList<List<Integer>>();

         List<Integer> res = new ArrayList<Integer>();

         if(root == null)

             return res;

         bfs(ret, root, );

         for(int i = ; i < ret.size(); ++i){

             res.add(ret.get(i).get(ret.get(i).size() - ));

         }

         return res;

     }

     public void bfs(List<List<Integer>> ret, TreeNode root, int dep){

         if(ret.size() <= dep){

             ret.add(new ArrayList<Integer>());

         }

         ret.get(dep).add(root.val);

         if(root.left != null)

             bfs(ret, root.left, dep + );

         if(root.right != null)

             bfs(ret, root.right, dep + );

     }

 }

巴特西

LeetCode OJ：Binary Tree Right Side View（右侧视角下的二叉树）

最新文章

热门文章