2415. Reverse Odd Levels of Binary Tree

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.

Complexity:

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

C++

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;
    }
};

JAVA

class Solution {
    public TreeNode reverseOddLevels(TreeNode root) {
        if(root == null)
            return root;
        
        Queue<TreeNode> queue = new LinkedList<>();
        int level = 0;

        queue.add(root);

        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){
                    queue.add(cur.left);
                    queue.add(cur.right);
                }
            }

            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;
    }
}

PYTHON

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;
    }
};