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]
Explanation:
Example 2:
Input: root = [1,2,3,4,null,null,null,5]
Output: [1,3,4,5]
Explanation:
Example 3:
Input: root = [1,null,3]
Output: [1,3]
Example 4:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Approach
- Do the simple BFS level order traversal and usually you end up at the right most side by the time you reach the end of traversing the level
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> view = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
q.add(root);
while(!q.isEmpty()) {
TreeNode right = null;
for(int i = q.size(); i > 0; i--) {
TreeNode n = q.poll();
if(n != null) {
right = n;
q.add(n.left);
q.add(n.right);
}
}
if (right != null)
view.add(right.val);
}
return view;
}
}