Binary Tree Right Side View

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]

.

Credits:

Special thanks to @amrsaqr for adding this problem and creating all test cases.
Breadth-First-Search
using a queue to store tree nodes of every level, and add the last tree node value of every level into result.

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
if (root == null)
return ret;
Queue<TreeNode> queue= new LinkedList<TreeNode>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
if (node.left != null)
queue.offer(node.left);
if (node.right != null)
queue.offer(node.right);

if (i == size - 1)
ret.add(node.val);
}
}

return ret;
}
}
Depth-First-Search

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> ret = new ArrayList<Integer>();
helper(ret, root, 1);
return ret;
}

private void helper(List<Integer> ret, TreeNode root, int level) {
if (root == null)
return;

if (level > ret.size())
ret.add(root.val);

helper(ret, root.right, level + 1);
helper(ret, root.left, level + 1);
}
}