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题目描述:/**
* 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) {
if(root == null){
return new List<int>();
}
var result = new List<int>();
RightView(new List<TreeNode>(){root}, result);
return result;
}
public void RightView(IList<TreeNode> nodes, IList<int> result)
{
if(!nodes.Any()){
return;
}
// add last node which is on the most right position
result.Add(nodes.Last().val);
// BFS
var children = new List<TreeNode>();
foreach(var n in nodes){
var nl = Children(n);
if(nl.Any()){
children.AddRange(nl);
}
}
RightView(children, result);
}
private IList<TreeNode> Children(TreeNode node){
if(node == null){
return new List<TreeNode>();
}
var list = new List<TreeNode>();
if(node.left != null){
list.Add(node.left);
}
if(node.right != null){
list.Add(node.right);
}
return list;
}
}版权声明:本文为博主原创文章,未经博主允许不得转载。
LeetCode -- Binary Tree Right Side View
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原文地址:http://blog.csdn.net/lan_liang/article/details/48575915