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,#,#,15,7},
3
/
9 20
/
15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
要注意读取的顺序，是折线。每次都应该按照不同的顺序。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int> > re;
vector<vector <TreeNode *> > t;
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
if( root != NULL)
{
vector<int> ve;
ve.push_back(root->val);
vector <TreeNode *> tr;
tr.push_back(root->right);
tr.push_back(root->left);
re.push_back(ve);
t.push_back(tr);
get(0);
}
return re;
}
void get(int n)
{
vector<int> ve;
vector <TreeNode *> tr;
if(n%2 == 0)
{
vector<TreeNode *> tt;
for(int i = 0 ; i < t[n].size();i++)
{
if(t[n][i] != NULL)
{
ve.push_back(t[n][i]->val);
tt.push_back(t[n][i]->right);
tt.push_back(t[n][i]->left);
}
}
for(int i = tt.size()-1;i>=0;i--)
tr.push_back(tt[i]);
}
else{
vector<TreeNode *> tt;
for(int i = 0 ; i < t[n].size();i++)
{
if(t[n][i] != NULL)
{
ve.push_back(t[n][i]->val);
tt.push_back(t[n][i]->left);
tt.push_back(t[n][i]->right);
}
}
for(int i = tt.size()-1;i>=0;i--)
tr.push_back(tt[i]);
}
if(ve.size()==0)return;
re.push_back(ve);
t.push_back(tr);
get(n+1);
return ;
}
};