LeetCode #199 - Binary Tree Right Side View

题目描述：

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]

.

层序遍历的每一层的最后一位即为所求。

class Solution {
public:
vector<int> rightSideView(TreeNode* root)
{
queue<pair<TreeNode*,int> > q;
vector<pair<int,int> > v;
if(root==NULL)
{
vector<int> result;
return result;
}
else if(root!=NULL)
{
pair<TreeNode*,int> p;
p.first=root;
p.second=0;
q.push(p);
while(!q.empty())
{
p=q.front();
pair<int,int> x;
x.first=p.first->val;
x.second=p.second;
v.push_back(x);
q.pop();
if(p.first->left!=NULL)
{
pair<TreeNode*,int> l;
l.first=p.first->left;
l.second=p.second+1;
q.push(l);
}
if(p.first->right!=NULL)
{
pair<TreeNode*,int> r;
r.first=p.first->right;
r.second=p.second+1;
q.push(r);
}
}
}
int n=v.size();
int m=v[n-1].second;
vector<vector<int> > result(m+1);
for(int i=0;i<n;i++)
{
result[v[i].second].push_back(v[i].first);
}
vector<int> Result(m+1);
for(int i=0;i<=m;i++)
{
Result[i]=result[i].back();
}
return Result;
}
};