Leetcode 2415 Reverse Odd Levels of Binary Tree Solution in Java | Hindi Coding Community

 

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.

Java Code :

class Solution {

public TreeNode reverseOddLevels(TreeNode root) {

traverse(root.left, root.right, 1);

return root;

}

public void traverse(TreeNode node1, TreeNode node2, int lvl) {

if (node1 == null || node2 == null) return;

if (lvl % 2 == 1){

int temp = node1.val;

node1.val = node2.val;

node2.val = temp;

}

traverse(node1.left, node2.right, lvl + 1);

traverse(node1.right, node2.left, lvl + 1);

}

}