LeetCode-199.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]

.
很简单的一道题，跟一层层遍历树的题型一样

/**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public IList<int> RightSideView(TreeNode root)
{
IList<int> res = new List<int>();
if (root == null)
return res;
Queue<TreeNode> q = new Queue<TreeNode>();
q.Enqueue(root);
int n;
TreeNode node;
while ((n=q.Count)>0)
{
res.Add(q.Peek().val);
for (int i = 0; i < n; i++)
{
node = q.Dequeue();
if (node.right != null)
q.Enqueue(node.right);
if (node.left != null)
q.Enqueue(node.left);
}
}
return res;
}
}

C++实现

#include <vector>
#include <queue>
using namespace std;
struct TreeNode
{
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

vector<int> rightSideView(TreeNode* root)
{
vector<int> res;
if (!root)
return res;
queue<TreeNode*> q;
q.push(root);
int n;
TreeNode* node;
while ((n = q.size()) > 0)
{
res.push_back(q.front()->val);
for (int i = 0; i < n; i++)
{
node = q.front();
q.pop();
if (node->right)
q.push(node->right);
if (node->left)
q.push(node->left);
}
}
return res;
}

递归解，参考https://leetcode.com/discuss/31348/my-simple-accepted-solution-java

public class Solution
{
public IList<int> RightSideView(TreeNode root)
{
IList<int> res = new List<int>();
Func(res, root, 0);
return res;
}

private void Func(IList<int> res, TreeNode root, int level)
{
if (root == null)
return;
if (level == res.Count)
res.Add(root.val);
Func(res,root.right,level + 1);
Func(res,root.left,level + 1);
}
}