103. Binary Tree Zigzag Level Order Traversal
Tags: ‘Stack’, ‘Tree’, ‘Breadth-first Search’
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
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new LinkedList<>();
if (root == null) return result;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> list = new LinkedList<>();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
list.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
if (result.size() % 2 == 1) {
Collections.reverse(list);
}
result.add(list);
}
return result;
}