LeetCode #103 - Binary Tree Zigzag Level Order Traversal

题目描述：

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:

Given binary tree

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

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

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

将层序遍历的奇数层的向量元素反过来即可。

class Solution {
public:
vector<vector<int>> zigzagLevelOrder(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=result;
for(int i=0;i<=m;i++)
{
if(i%2==1)
{
for(int j=0;j<result[i].size();j++)
{
Result[i][j]=result[i][result[i].size()-1-j];
}
}
}
return Result;
}
};