103.leetcode Binary Tree Zigzag Level Order Traversal(medium)[二叉树 栈]

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]
]

具体解题思路：要采用之字形按层次遍历树，这里采用了两个栈来完成，由于遍历的时候是交叉从左到右或者从右到左的顺序，所以这里采用了flag来判断当前是应该按照从左到右还是从右到左的顺序。st1保存的是需要从右到左来遍历下一层的节点，st2保存的是需要从左到右来遍历下一层的节点。因此判断的依据就是当st1和st2都为空时整棵树就遍历完成。

class Solution {
public:

vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
//vector<int> temp;
vector<vector<int> > result;
if(root == NULL) return result;
stack<TreeNode*> st1;
stack<TreeNode*> st2;
int flag = 0;
st1.push(root);
while(st1.size()!=0 || st2.size() !=0)
{
if(flag == 0 && st1.size()>0)
{
vector<int> temps;
while(st1.size()>0)
{
TreeNode* temp = st1.top();
if(temp->left != NULL)
st2.push(temp->left);
if(temp->right != NULL)
st2.push(temp->right);
st1.pop();
temps.push_back(temp->val);
}
result.push_back(temps);
flag = 1;
}

if(flag == 1 && st2.size()>0)
{
vector<int> temps;
while(st2.size()>0)
{
TreeNode* temp = st2.top();
if(temp->right != NULL)
st1.push(temp->right);
if(temp->left != NULL)
st1.push(temp->left);
st2.pop();
temps.push_back(temp->val);
}
result.push_back(temps);
flag = 0;
}
}
return result;
}
};