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.

Python3JavaC++GoTypeScriptRust

# 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.
 * 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) {
                    t.add(node);
                }
                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 =