[LeetCode] Binary Tree Zigzag Level Order Traversal

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]
]

用BFS然后根据层数来判断是否要反转列表

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

struct Node
{
TreeNode *node;
int level;
Node(){}
Node(TreeNode *n, int l):node(n), level(l){}
};

class Solution {
private:
vector<vector<int> > ret;
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
ret.clear();

if (root == NULL)
return ret;

queue<Node> q;

q.push(Node(root, 0));

int curLevel = -1;

vector<int> a;

while(!q.empty())
{
Node node = q.front();

if (node.node->left)
q.push(Node(node.node->left, node.level + 1));

if (node.node->right)
q.push(Node(node.node->right, node.level + 1));

if (curLevel != node.level)
{
if (curLevel != -1)
{
if (curLevel % 2 == 1)
reverse(a.begin(), a.end());

ret.push_back(a);
}
a.clear();
curLevel = node.level;
}
a.push_back(node.node->val);
q.pop();
}

if (curLevel % 2 == 1)
reverse(a.begin(), a.end());

ret.push_back(a);

return ret;
}
};