难度:中等
给定一个二叉树,返回其节点值的锯齿形层次遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
例如:
给定二叉树 [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
返回锯齿形层次遍历如下:
[ [3], [20,9], [15,7] ]
题目分析:
使用两个队列,但是偶数队列存储的时候从左到右存储,奇数队列存储的时候从右到左。
参考代码:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int> > res;
if(root == NULL)
return res;
stack<TreeNode*> odd_stack;
stack<TreeNode*> even_stack;
vector<int> temp;
TreeNode* pMove = root;
odd_stack.push(root);
while(!odd_stack.empty() || !even_stack.empty())
{
while(!odd_stack.empty())
{
pMove = odd_stack.top();
odd_stack.pop();
temp.push_back(pMove->val);
//偶数队列从左到右存储
if(pMove->left)
even_stack.push(pMove->left);
if(pMove->right)
even_stack.push(pMove->right);
}
if(!temp.empty())
{
res.push_back(temp);
temp.clear();
}
while(!even_stack.empty())
{
pMove = even_stack.top();
even_stack.pop();
temp.push_back(pMove->val);
//奇数队列从左到右存储
if(pMove->right)
odd_stack.push(pMove->right);
if(pMove->left)
odd_stack.push(pMove->left);
}
if(!temp.empty())
{
res.push_back(temp);
temp.clear();
}
}
return res;
}
};