# 2415. Reverse Odd Levels of Binary Tree

## Description

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.

## Solutions

Solution 1: BFS

We can use the Breadth-First Search (BFS) method, using a queue $q$ to store the nodes of each level, and a variable $i$ to record the current level. If $i$ is odd, we reverse the values of the nodes at the current level.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

• /**
* 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) {
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
for (int i = 0; !q.isEmpty(); ++i) {
List<TreeNode> t = new ArrayList<>();
for (int k = q.size(); k > 0; --k) {
var node = q.poll();
if (i % 2 == 1) {
}
if (node.left != null) {
q.offer(node.left);
q.offer(node.right);
}
}
for (int l = 0, r = t.size() - 1; l < r; ++l, --r) {
var x = t.get(l).val;
t.get(l).val = t.get(r).val;
t.get(r).val = x;
}
}
return root;
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* reverseOddLevels(TreeNode* root) {
queue<TreeNode*> q{ {root} };
for (int i = 0; q.size(); ++i) {
vector<TreeNode*> t;
for (int k = q.size(); k; --k) {
TreeNode* node = q.front();
q.pop();
if (i & 1) {
t.push_back(node);
}
if (node->left) {
q.push(node->left);
q.push(node->right);
}
}
for (int l = 0, r = t.size() - 1; l < r; ++l, --r) {
swap(t[l]->val, t[r]->val);
}
}
return root;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def reverseOddLevels(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
q = deque([root])
i = 0
while q:
if i & 1:
l, r = 0, len(q) - 1
while l < r:
q[l].val, q[r].val = q[r].val, q[l].val
l, r = l + 1, r - 1
for _ in range(len(q)):
node = q.popleft()
if node.left:
q.append(node.left)
q.append(node.right)
i += 1
return root


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func reverseOddLevels(root *TreeNode) *TreeNode {
q := []*TreeNode{root}
for i := 0; len(q) > 0; i++ {
t := []*TreeNode{}
for k := len(q); k > 0; k-- {
node := q[0]
q = q[1:]
if i%2 == 1 {
t = append(t, node)
}
if node.Left != nil {
q = append(q, node.Left)
q = append(q, node.Right)
}
}
for l, r := 0, len(t)-1; l < r; l, r = l+1, r-1 {
t[l].Val, t[r].Val = t[r].Val, t[l].Val
}
}
return root
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function reverseOddLevels(root: TreeNode | null): TreeNode | null {
const q: TreeNode[] = [root];
for (let i = 0; q.length > 0; ++i) {
if (i % 2) {
for (let l = 0, r = q.length - 1; l < r; ++l, --r) {
[q[l].val, q[r].val] = [q[r].val, q[l].val];
}
}
const nq: TreeNode[] = [];
for (const { left, right } of q) {
if (left) {
nq.push(left);
nq.push(right);
}
}
q.splice(0, q.length, ...nq);
}
return root;
}


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
fn create_tree(vals: &Vec<Vec<i32>>, i: usize, j: usize) -> Option<Rc<RefCell<TreeNode>>> {
if i == vals.len() {
return None;
}
Some(
Rc::new(
RefCell::new(TreeNode {
val: vals[i][j],
left: Self::create_tree(vals, i + 1, j * 2),
right: Self::create_tree(vals, i + 1, j * 2 + 1),
})
)
)
}

pub fn reverse_odd_levels(
root: Option<Rc<RefCell<TreeNode>>>
) -> Option<Rc<RefCell<TreeNode>>> {
let mut queue = VecDeque::new();
queue.push_back(root);
let mut d = 0;
let mut vals = Vec::new();
while !queue.is_empty() {
let mut val = Vec::new();
for _ in 0..queue.len() {
let mut node = queue.pop_front().unwrap();
let mut node = node.as_mut().unwrap().borrow_mut();
val.push(node.val);
if node.left.is_some() {
queue.push_back(node.left.take());
}
if node.right.is_some() {
queue.push_back(node.right.take());
}
}
if d % 2 == 1 {
val.reverse();
}
vals.push(val);
d += 1;
}
Self::create_tree(&vals, 0, 0)
}
}