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Formatted question description: https://leetcode.ca/all/2415.html

# 2415. Reverse Odd Levels of Binary Tree

• Difficulty: Medium.
• Related Topics: .
• Similar Questions: Invert Binary Tree.

## Problem

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 (Java, C++, Python)

• /**
* 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) {
if(root == null)
return root;

int level = 0;

while (!queue.isEmpty()) {
int size = queue.size();
level++;

TreeNode[] arr = new TreeNode[size];

for (int i = 0; i < size; i++) {
TreeNode cur = queue.poll();
arr[i] = cur;

if(cur.left != null){
}
}

if (level % 2 == 0) {
int left = 0;
int right = size - 1;

while (left < right) {
int leftVal = arr[left].val;
arr[left].val = arr[right].val;
arr[right].val = leftVal;

left++;
right--;
}
}
}

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:
t = []
for _ in range(len(q)):
node = q.popleft()
if i & 1:
t.append(node)
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
if t:
j, k = 0, len(t) - 1
while j < k:
t[j].val, t[k].val = t[k].val, t[j].val
j, k = j + 1, k - 1
i += 1
return root

############

# 2415. Reverse Odd Levels of Binary Tree
# https://leetcode.com/problems/reverse-odd-levels-of-binary-tree/

# 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]:
res = []

dq = deque([root])
level = 0

while dq:
n = len(dq)
curr = []

for _ in range(n):
node = dq.popleft()
curr.append(node.val)

for child in filter(None, (node.left, node.right)):
dq.append(child)

if level % 2 == 1:
curr.reverse()

res += curr
level += 1

n = len(res)

def go(i):
root = None

if i < n:
root = TreeNode(res[i])

# insert left child
root.left = go(2 * i + 1)

# insert right child
root.right = go(2 * i + 2)

return root

return go(0)


• /**
* 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} };
int i = 0;
vector<TreeNode*> t;
while (!q.empty()) {
t.clear();
for (int n = q.size(); n; --n) {
TreeNode* node = q.front();
q.pop();
if (i & 1) {
t.push_back(node);
}
if (node->left) {
q.push(node->left);
}
if (node->right) {
q.push(node->right);
}
}
if (t.size()) {
int j = 0, k = t.size() - 1;
for (; j < k; ++j, --k) {
int v = t[j]->val;
t[j]->val = t[k]->val;
t[k]->val = v;
}
}
++i;
}
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}
i := 0
for len(q) > 0 {
t := []*TreeNode{}
for n := len(q); n > 0; n-- {
node := q[0]
q = q[1:]
if i%2 == 1 {
t = append(t, node)
}
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
}
if len(t) > 0 {
j, k := 0, len(t)-1
for ; j < k; j, k = j+1, k-1 {
v := t[j].Val
t[j].Val = t[k].Val
t[k].Val = v
}
}
i++
}
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 queue = [root];
let d = 0;
while (queue.length !== 0) {
const n = queue.length;
const t: TreeNode[] = [];
for (let i = 0; i < n; i++) {
const node = queue.shift();
if (d % 2 == 1) {
t.push(node);
}
node.left && queue.push(node.left);
node.right && queue.push(node.right);
}
const m = t.length;
for (let i = 0; i < m >> 1; i++) {
[t[i].val, t[m - 1 - i].val] = [t[m - 1 - i].val, t[i].val];
}
d++;
}
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)
}
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).