103. Binary Tree Zigzag Level Order Traversal
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:
- The number of nodes in the tree is in the range [0, 2000].
- -100 <= Node.val <= 100
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> listNode = new LinkedList<>();
Stack<TreeNode> stack1 = new Stack<>();
Stack<TreeNode> stack2 = new Stack<>();
if(root == null){
return listNode;
}
int flag = 0;
stack1.push(root);
while(!stack1.isEmpty()||!stack2.isEmpty()){
List<Integer> subList = new LinkedList<>();
if(flag%2==0){
while(!stack1.isEmpty()){
TreeNode temp = stack1.pop();
if(temp!=null){
subList.add(temp.val);
stack2.push(temp.left);
stack2.push(temp.right);
}
}
}else{
while(!stack2.isEmpty()){
TreeNode temp = stack2.pop();
if(temp!=null){
subList.add(temp.val);
stack1.push(temp.right);
stack1.push(temp.left);
}
}
}
if(!subList.isEmpty()){
listNode.add(subList);
}
flag++;
}
return listNode;
}
}