【原创】leetCodeOj --- Binary Tree Right Side View 解题报告

二连水

 

题目地址：

https://leetcode.com/problems/binary-tree-right-side-view/

 

题目内容：

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].

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

 

题目解析：

乍一看，酷炫异常，实际上单纯得让人想哭。

让你找到二叉树每层的最后一个元素。

怎么早？

BFS一层，队列的最后一个节点。

 

具体代码：

/**
 * 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) {
        Deque<TreeNode> bfs = new LinkedList<TreeNode>();
        List<Integer> result = new ArrayList<Integer>();
        if (root == null) {
            return result;
        }
        bfs.add(root);
        while (!bfs.isEmpty()) {
            TreeNode last = bfs.getLast();
            result.add(last.val);
            TreeNode first = bfs.getFirst();
            while (first != last) {
                if (first.left != null) {
                    bfs.add(first.left);
                }
                if (first.right != null) {
                    bfs.add(first.right);
                }
                bfs.pop();
                first = bfs.getFirst();
            }
            // handle last
            if (last.left != null) {
                bfs.add(last.left);
            }
            if (last.right != null) {
                bfs.add(last.right);
            }
            bfs.pop();
        }
        return result;
    }
}