leetcode---Binary Tree Zigzag Level Order Traversal---层次遍历

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]

]

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

struct Node
{
int level;
TreeNode *tnode;

Node(){}

Node(TreeNode *n, int l)
{
tnode = n;
level = l;
}
};

vector<vector<int>> ans;

class Solution
{
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root)
{
ans.clear();
vector<int> tmp;
if(root == NULL)
return ans;
queue<Node> q;
q.push(Node(root, 0));
int curLevel = -1;
while(!q.empty())
{
Node node = q.front();
q.pop();
if(node.tnode->left)
q.push(Node(node.tnode->left, node.level+1));
if(node.tnode->right)
q.push(Node(node.tnode->right, node.level+1));
if(curLevel != node.level)
{
if(curLevel != -1)
{
if(curLevel & 1 == 1)
{
reverse(tmp.begin(), tmp.end());
}
ans.push_back(tmp);
}
tmp.clear();
curLevel = node.level;
}
tmp.push_back(node.tnode->val);
}
if(curLevel & 1 == 1)
reverse(tmp.begin(), tmp.end());
ans.push_back(tmp);
return ans;
}
};