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。