Given the root of 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.
Example 1:
Input: root = [1,2,3,null,5,null,4] Output: [1,3,4] Example 2:
Input: root = [1,null,3] Output: [1,3] Example 3:
Input: root = [] Output: []
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> res = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
if(root == null) return res;
queue.offer(root);
while(!queue.isEmpty()){
int size = queue.size();
for(int i =0; i<size; i++){
TreeNode curr = queue.poll();
if(i==0) res.add(curr.val);
if(curr.right != null) queue.offer(curr.right);
if(curr.left != null) queue.offer(curr.left);
}
}
return res;
}
}
Time Complexity: O(N) Space Complexity: O(N)
和 102. Binary Tree Level Order Traversal 做比較