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.
Example:
Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:
1 <---
/
2 3 <---
5 4 <---
Approach #1: C++. [recursive]
/**
* 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) {
vector<int> ans;
helper(root, ans, 1);
return ans;
}
private:
void helper(TreeNode* root, vector<int>& ans, int level) {
if (root == NULL) return;
if (ans.size() < level) ans.push_back(root->val);
helper(root->right, ans, level+1);
helper(root->left, ans, level+1);
}
};
Approach #2: Java. [bfs + queue]
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
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 cur = queue.poll();
if (i == 0) res.add(cur.val);
if (cur.right != null) queue.offer(cur.right);
if (cur.left != null) queue.offer(cur.left);
}
}
return res;
}
}
Approach #3: Python. [Iterator]
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def rightSideView(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
view = []
if root:
level = [root]
while level:
view += level[-1].val,
level = [kid for node in level for kid in (node.left, node.right) if kid]
return view