TODO: 103. Binary Tree Zigzag Level Order Traversal
Problem
Intuition
Time Complexity
O(V+E)
-> Standard BFS time complexity
Space Complexity
O(V)
-> Standard BFS space complexity
Solution
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
// TC: O(V+E)
List<List<Integer>> result = new ArrayList<>();
if(root == null) {
return result;
}
// Performing standard BFS
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
// Used as flag to understand if elements in level have to be ordered
// from left to right or vice-verca
int level = 0;
while(!q.isEmpty()) {
int n = q.size();
List<Integer> current = new LinkedList<>();
for(int i=0;i<n;i++) {
TreeNode node = q.poll();
// This toggle controls, in which order the elements will be
// inserted to current
if(level % 2 == 0) {
current.addLast(node.val);
} else {
current.addFirst(node.val);
}
if(node.left != null) q.offer(node.left);
if(node.right != null) q.offer(node.right);
}
result.add(current);
level++;
}
return result;
}
}
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