【leetcode】Binary Tree Zigzag Level Order Traversal （middle）

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]
]

思路：由于需要排成之字形，所以用一个栈来存储当前行的内容，另一个栈来存储下一行的内容。

以根为第0层，那么偶数层都应该从左向右输出。那么每次遇到偶数层，压入下一层奇数层时就按从左到右的顺序，这样弹栈时就是从右到左。

纠结了一会儿，AC了。

class Solution {
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
vector<vector<int>> ans;
if(root == NULL)
{
return ans;
}
int level = 0; //记录当前层号
vector<TreeNode *> curlevel;
curlevel.push_back(root);
while(!curlevel.empty())
{
vector<int> partans;
vector<TreeNode *> nextlevel;
if(level % 2 == 0) //偶数层，根是第0层
{
while(!curlevel.empty()) //本层不为空
{
if(curlevel.back()->left != NULL) //先压入左边，再压入右边
{
nextlevel.push_back(curlevel.back()->left);
}
if(curlevel.back()->right != NULL)
{
nextlevel.push_back(curlevel.back()->right);
}
partans.push_back(curlevel.back()->val);
curlevel.pop_back();
}
ans.push_back(partans);
curlevel = nextlevel;
}
else
{
while(!curlevel.empty()) //本层不为空
{
if(curlevel.back()->right != NULL) //先压入右边，再压入左边
{
nextlevel.push_back(curlevel.back()->right);
}
if(curlevel.back()->left != NULL)
{
nextlevel.push_back(curlevel.back()->left);
}
partans.push_back(curlevel.back()->val);
curlevel.pop_back();
}
ans.push_back(partans);
curlevel = nextlevel;
}
level++;
}
return ans;
}

void createTree(TreeNode * &root)
{
int n;
cin >> n;
if(n != 0)
{
root = new TreeNode(n);
createTree(root->left);
createTree(root->right);
}
}
};

看看别人的答案。思路不一样。

class Solution {
vector<vector<int> > result;
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {

if(root!=NULL)
{
traverse(root, 0);
}

for(int i=1;i<result.size();i+=2)
{
vector<int>* v = &result[i];
std:reverse(v->begin(), v->end());
}
return result;
}

void traverse(TreeNode* node, int level)
{
if(node == NULL) return;

vector<int>* row = getRow(level);
row->push_back(node->val);

traverse(node->left, level+1);
traverse(node->right, level+1);
}

vector<int>* getRow(int level)
{
if(result.size()<=level)
{
vector<int> newRow;
result.push_back(newRow);
}
return &result[level];
}
};