199. Binary Tree Right Side View(BFS).java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
  /*
 1. BFS时，先遍历右子树、后左子树，把每层的第一个加到res里，（如果res.size < level时，add，或者跳过）
 2. BFS时，先左子树后右子树，每一层对应一个值，前面的会把后面的不断覆盖。
 */
public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null) return res;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while(!queue.isEmpty()) {
            TreeNode cur = queue.peek();
            int curSize = queue.size();
            for(int i = 0; i < curSize; i++) {
                cur = queue.poll();
                if(cur.left != null) {
                    queue.offer(cur.left);
                }
                if(cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            res.add(cur.val);
        }
        return res;
    }
}

/*
[]
[1,2,3,null,5,null,4]
*/

199. Binary Tree Right Side View(DFS).java

/**
 * 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> result = new ArrayList<>();
        calc(root, result,0);
        return result;
    }
    public void calc(TreeNode root, List<Integer> result, int depth){
        if(root==null) return;
        if(depth==result.size()){ result.add(root.val); } //result depth basically shows for every depth level, there should be one node entered i.e. when result size is 2 there should be 2 elements when it's 3 there should be 3 n so on
        calc(root.right, result, depth+1);
        calc(root.left, result, depth+1);
    }
}