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]
]
Solution:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> ans = new ArrayList<>();
if(root==null) return ans;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
boolean isForward = true;
while(!queue.isEmpty()){
Queue<TreeNode> tmp = new LinkedList<>();
List<Integer> level = new ArrayList<>();
while(!queue.isEmpty()){
TreeNode tn = queue.poll();
if(isForward){
level.add(tn.val);
} else {
level.add(0, tn.val);
}
if(tn.left!=null) tmp.offer(tn.left);
if(tn.right!=null) tmp.offer(tn.right);
}
queue = tmp;
ans.add(level);
isForward = !isForward;
}
return ans;
}
}