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]
]

蛇形遍历二叉树，这里可以将二叉树层序遍历，然后把奇数层的反转，就可以。

这里考虑可不可以直接做呢？蛇形遍历上一层的尾是下一层的头，这正好符合stack的性质，所以用两个stack循环使用就可以解决这个问题

class Solution {
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
stack<TreeNode *> stack1, stack2;
vector<vector<int> > result;
if (NULL == root) return result;
stack1.push(root);
while (!stack1.empty() || !stack2.empty()){
vector<int> v;
while (!stack1.empty()){
TreeNode *p = stack1.top();
v.push_back(p->val);
if (p->left) stack2.push(p->left);
if (p->right) stack2.push(p->right);
stack1.pop();

}
if (!v.empty())
{
result.push_back(v);
v.clear();
}
while (!stack2.empty()){
TreeNode *p = stack2.top();
v.push_back(p->val);
if (p->right) stack1.push(p->right);
if (p->left) stack1.push(p->left);
stack2.pop();

}
if (!v.empty())
{
result.push_back(v);
v.clear();
}
}
return result;
}
};