Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).
Example 1:
Input: root = [3,9,20,null,null,15,7] Output: [[3],[20,9],[15,7]]
Example 2:
Input: root = [1] Output: [[1]]
Example 3:
Input: root = [] Output: []
Constraints:
[0, 2000].-100 <= Node.val <= 100
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
if(root==null){
return ans ;
}
List<Integer> currAns = new ArrayList<>();
Stack<TreeNode> main = new Stack<>();
Stack<TreeNode> helper = new Stack<>();
int level = 0 ;
main.push(root);
while(main.size()>0){
TreeNode currNode = main.pop();
currAns.add(currNode.val);
if(level == 0 ){
if(currNode.left !=null){
helper.push(currNode.left);
}
if(currNode.right !=null){
helper.push(currNode.right);
}
}else{
if(currNode.right !=null){
helper.push(currNode.right);
}
if(currNode.left !=null){
helper.push(currNode.left);
}
}
if(main.size()==0){
ans.add(currAns);
currAns = new ArrayList<>();
level = 1-level ;
main = helper ;
helper = new Stack<>();
}
}
return ans ;
}
}
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