199. Binary Tree Right Side View

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LeetCode

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Problem

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       <---

Code

/**
 * 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;
        helper(res, root, 0);
        return res;
    }

    private void helper(List<Integer> res, TreeNode root, int level){
        if(root == null) return;
        // 保存当前树的等级
        // 每个节点要能加入到结果的前提是必须属于这个等级的节点
        // 但是是从右边看，因此要优先看看右子树有没有结果
        if(res.size() == level){
            res.add(root.val);
        }
        // 优先查找右子树
        helper(res, root.right, level + 1);
        helper(res, root.left, level + 1);
    }
}

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        // reverse level traversal
        List<Integer> result = new ArrayList();
        Queue<TreeNode> queue = new LinkedList();
        if (root == null) return result;

        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode cur = queue.poll();
                // 第一个元素是这一层最右边的元素
                if (i == 0) result.add(cur.val);
                // 相当于从右往左加入
                if (cur.right != null) queue.offer(cur.right);
                if (cur.left != null) queue.offer(cur.left);
            }
        }
        return result;
    }
}