leetcode--199. Binary Tree Right Side View

1、问题描述

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.

For example:
Given the following binary tree,

   1            <---
 /   
2     3         <---
      
  5     4       <---

You should return [1, 3, 4].

2、边界条件：root==null

3、思路：从右边看，是看到了每一行的最右边一个，也就是每一行从左往右遍历的最后一个。或者从右往左遍历的第一个。

可以按照上面两种实现。

4、代码实现

方法一：从左边往右遍历：

/**
 * 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> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int curLevelNum = 1;
        while (!queue.isEmpty()) {
            TreeNode top = new TreeNode(0);
            int nextLevelNum = 0;
            while (curLevelNum-- > 0) {
                top = queue.poll();
                if (top.left != null) {
                    queue.add(top.left);
                    nextLevelNum++;
                }
                if (top.right != null) {
                    queue.add(top.right);
                    nextLevelNum++;
                }
            }
            result.add(top.val);
            curLevelNum = nextLevelNum;
        }
        return result;
    }
}

方法二：从右往左遍历

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        int curLevelNum = 1;
        while (!queue.isEmpty()) {
            result.add(queue.peek().val);
            int nextLevelNum = 0;
            while (curLevelNum-- > 0) {
                TreeNode top = queue.poll();
                if (top.right != null) {
                    queue.add(top.right);
                    nextLevelNum++;
                }
                if (top.left != null) {
                    queue.add(top.left);
                    nextLevelNum++;
                }
            }
            curLevelNum = nextLevelNum;
        }
        return result;
    }
}