Question
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].
Solution
User Depth First Search(DFS)
Traverse the binary tree. For a node, first visit its right child, then visit its left child. If it’s the first visit to the layer, add the value of the node to the result.
Code
Here is a sample solution.
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> rightSideView = new ArrayList();
collectRightSideView(root, rightSideView, 1);
return rightSideView;
}
void collectRightSideView(TreeNode root, List<Integer> rightSideView, int level) {
if(root == null) return;
if(level > rightSideView.size()) {
rightSideView.add(root.val);
}
collectRightSideView(root.right, rightSideView, level+1);
collectRightSideView(root.left, rightSideView, level+1);
}
}
Performance
Using DFS to traverse a tree needs O(n) time complexity and O(n) space complexity.