Binary Tree Right Side View

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

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].

解题思路：

这题比较简单了，层次遍历，每层输出最后一个node就可以了。

/**

 * Definition for binary tree

 * public class TreeNode {

 *     int val;

 *     TreeNode left;

 *     TreeNode right;

 *     TreeNode(int x) { val = x; }

 * }

 */

public class Solution {

    public List<Integer> rightSideView(TreeNode root) {

        List<Integer> result = new ArrayList<Integer>();

        if(root == null) {

            return result;

        }

        Queue<TreeNode> queue = new LinkedList<TreeNode>();

        queue.offer(root);

        

        while(queue.size() > 0) {

            int size = queue.size();

            while(size > 0) {

                TreeNode current = queue.poll();

                if(current.left != null) {

                    queue.offer(current.left);

                }

                if(current.right != null) {

                    queue.offer(current.right);

                }

                if(size == 1) {

                    result.add(current.val);

                }

                size--;

            }

        }

        return result;

    }

}

update 2015/06/13:

看到一个精妙的递归写法。精髓就在于两点。

https://leetcode.com/discuss/31348/my-simple-accepted-solution-java

1. 遇到当前深度==res大小的第一个节点，加入结果。

2. 始终首先处理右子树，再处理左子树。

/**

 * 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) {

        List<Integer> res = new ArrayList<Integer>();

        helper(root, res, 0);

        return res;

    }

    

    public void helper(TreeNode root, List<Integer> res, int level) {

        if(root == null) {

            return;

        }

        if(level == res.size()) {

            res.add(root.val);

        }

        helper(root.right, res, level + 1);

        helper(root.left, res, level + 1);

    }

}

原文里也提到，这么做的时间复杂度是O(n)。因为在任何情况下，他都要遍历完所有的节点。