199. Binary Tree Right Side View - Medium

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       <---

 

M1: BFS

level order traverse,，每次遍历完一层后，只把这一层最右的数字加入res中

time: O(n), space: O(N)　　-- N: most number of nodes in one level

/**
 * 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> q = new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()) {
            List<Integer> tmp = new ArrayList<>();
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode t = q.poll();
                tmp.add(t.val);
                if(t.left != null) {
                    q.offer(t.left);
                }
                if(t.right != null) {
                    q.offer(t.right);
                }
            }
            res.add(tmp.get(tmp.size() - 1));
        }
        
        return res;
    }
}

 

M2: DFS

ref: https://leetcode.com/problems/binary-tree-right-side-view/discuss/56012/My-simple-accepted-solution(JAVA)

time: O(n), space: O(n)

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