199.Binary Tree Right Side View

Posted on | In

Tree, Medium

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.

Example:

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Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:
   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

Answer

Iterative
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/**
 * 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<>();
        if(root == null)
            return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while(!queue.isEmpty()){
            int size = queue.size();
            for(int i = 0; i < size; i++){
                TreeNode node = queue.poll();
                if(i == 0)
                    res.add(node.val);
                if(node.right != null)
                    queue.offer(node.right);
                if(node.left != null)
                    queue.offer(node.left);
            }
        }
        
        return res;
    }
}

Time complexity: O(n)

DFS, Recursive
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/**
 * 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<>();
        if(root == null)
            return res;
        recursive(res, root, 0);
        return res;
    }
    public void recursive(List<Integer> res, TreeNode node, int level){
        if(node == null)
            return;
        if(res.size() == level){
            res.add(node.val);
        }
        recursive(res, node.right, level + 1);
        recursive(res, node.left, level + 1);
    }
}

Time complexity: O(n), 3ms