199. Binary Tree Right Side View

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].

 

其实就是层序遍历 的每层的最后一个元素！！！！

 

 

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
import queue
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        q = queue.Queue()
        res = []
        if root == None:
            return []
        q.put(root)
        while q.qsize():
            sz = q.qsize()
            for i in range(sz):
                top = q.get()
                if i == sz -1:
                    res.append(top.val)
                if top.left : q.put(top.left)
                if top.right : q.put(top.right)
        return res

 

 

 

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> res ;
        if(root == nullptr) return res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()) {
            int cnt = q.size();
            for(int i = 0 ; i < cnt ; ++i) {
                auto node = q.front();q.pop();
                if(i == cnt -1) res.push_back(node->val);
                if(node->left != nullptr) q.push(node->left);
                if(node->right!= nullptr) q.push(node->right);
            }
        }
        return res;
    }
};

 

class Solution {
    
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        if(root==null) return res;
        queue.offer(root);
        int level_num = 1;
        while (!queue.isEmpty()) {
            level_num = queue.size();
            for(int i = 0; i < level_num; i++){
                TreeNode node = queue.poll();
                if(i==level_num-1)
                    res.add(node.val);
                if(node.left != null) queue.offer(node.left);
                if(node.right != null) queue.offer(node.right);
                
            }
        }
        return res;
    }
}