描述:
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].
思路:
按层序遍历的思路,每次只保存该层的最后一个元素即可。
代码:
public List<Integer> rightSideView(TreeNode root) {
List<Integer>listReturn=new ArrayList<Integer>();
if(root==null)
return listReturn;
List<TreeNode>list=new ArrayList<TreeNode>();
List<TreeNode>temp=new ArrayList<TreeNode>();
list.add(root);
TreeNode node=null;
listReturn.add(root.val);
while(list.size()!=0)
{
for(int i=0;i<list.size();i++)
{
node=list.get(i);
if(node.left!=null)
temp.add(node.left);
if(node.right!=null)
temp.add(node.right);
}
if(temp.size()>0)
listReturn.add(temp.get(temp.size()-1).val);
list.clear();
list=temp;
temp=new ArrayList<TreeNode>();
}
return listReturn;
}
结果:
leetocde_Binary Tree Right Side View
原文地址:http://blog.csdn.net/mnmlist/article/details/44945941