LeetCode75-199-BT-Right-Side-View

🧠 Summary of Thoughts

Idea

Use level-order traversal (BFS) to process the tree level by level. For each level, record the last node because it represents the node visible from the right side.

Steps

• Use a queue to perform BFS.

• For each level, get the number of nodes: size = queue.size().

• Traverse all nodes in that level.

• When visiting the last node of the level (i == size - 1), add its value to the result.

• Push the node’s left and right children into the queue.

When to use BFS?

1. Level by Level
2. shortest path
3. spread from one spot

✅ Code（Java）

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 == size - 1) {
                    res.add(node.val);
                }
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
        }
        return res;
    }
}

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        dfs(root, 0, res);
        return res;
    }
    private void dfs (TreeNode node, int depth, List<Integer> res) {
        if (node == null) return;
        // If it is the first time to this level, then add the rightmost node.val to the res
        if (depth == res.size()) {
            res.add(node.val);
        }
        dfs(node.right, depth + 1, res);
        dfs(node.left, depth + 1, res);
    }
}