2493. Reverse Odd Levels Of Binary Tree¶

2493. Reverse Odd Levels of Binary Tree

Medium

Given the root of a perfect binary tree, reverse the node values at each odd level of the tree.

For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18], then it should become [18,29,11,7,4,3,1,2].

Return the root of the reversed tree.

A binary tree is perfect if all parent nodes have two children and all leaves are on the same level.

The level of a node is the number of edges along the path between it and the root node.

Example 1:

Input: root = [2,3,5,8,13,21,34]
Output: [2,5,3,8,13,21,34]
Explanation: 
The tree has only one odd level.
The nodes at level 1 are 3, 5 respectively, which are reversed and become 5, 3.

Example 2:

Input: root = [7,13,11]
Output: [7,11,13]
Explanation: 
The nodes at level 1 are 13, 11, which are reversed and become 11, 13.

Example 3:

Input: root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
Output: [0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
Explanation: 
The odd levels have non-zero values.
The nodes at level 1 were 1, 2, and are 2, 1 after the reversal.
The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1 after the reversal.

Constraints:

The number of nodes in the tree is in the range [1, 2¹⁴].
0 <= Node.val <= 10⁵
root is a perfect binary tree.

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 TreeNode reverseOddLevels(TreeNode root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while (q.size() > 0) {
            int len = q.size();
            ArrayList<Integer> temp = new ArrayList<>();
            for (int i = 0; i < len; i++) {
                if (q.peek().left != null) q.offer(q.peek().left);
                if (q.peek().right != null) q.offer(q.peek().right);
                temp.add(q.poll().val);
            }
            res.add(new ArrayList<>(temp));
        }
        for (int i = 0; i < res.size(); i++) if (i % 2 == 1) Collections.reverse(res.get(i));
        ArrayList<Integer> nodes = new ArrayList<>();
        for (int i = 0; i < res.size(); i++) {
            for (int ele : res.get(i)) nodes.add(ele);
        }
        return BuildTree(nodes);
    }
    private TreeNode BuildTree(ArrayList<Integer> nodes) {
        int n = nodes.size();
        if (n == 0) return null;
        TreeNode root = new TreeNode(nodes.get(0));
        int current_index = 1;
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while (q.size() > 0) {
            TreeNode current_node = q.poll();
            int left = -1, right = -1;
            if (current_index < n) left = nodes.get(current_index++);
            if (current_index < n) right = nodes.get(current_index++);
            if (left == -1) current_node.left = null;
            if (right == -1) current_node.right = null;
            if (left != -1) {
                current_node.left = new TreeNode(left);
                q.offer(current_node.left);
            }
            if (right != -1) {
                current_node.right = new TreeNode(right);
                q.offer(current_node.right);
            }
        }
        return root;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Explanation¶

[Add detailed explanation here]