LeetCode #107 - Binary Tree Level Order Traversal II

题目描述：

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:

Given binary tree

[3,9,20,null,null,15,7]

,

3
/ \
9  20
/  \
15   7

return its bottom-up level order traversal as:

[
[15,7],
[9,20],
[3]
]

从上到下得到的层序遍历反过来即可。

class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root)
{
queue<pair<TreeNode*,int> > q;
vector<pair<int,int> > v;
if(root==NULL)
{
vector<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<vector<int> > Result(m+1);
for(int i=0,j=m;i<=m&&j>=0;i++,j--)
{
Result[i]=result[j];
}
return Result;
}
};