LeetCode 2415: Reverse Odd Levels of Binary Tree
LeetCode 2415 Solution Explanation
Explanation
To solve this problem, we can perform a depth-first traversal of the binary tree and reverse the node values at each odd level. We can keep track of the current level while traversing the tree and reverse the values when we encounter an odd level.
- Perform a depth-first traversal of the binary tree, keeping track of the current level.
- When visiting a node, check if the level is odd. If it is odd, reverse the values of the nodes at that level.
- Recur for the left and right subtrees, incrementing the level as we go deeper.
- Return the modified root of the binary tree.
Time Complexity: O(N) where N is the number of nodes in the binary tree. Space Complexity: O(H) where H is the height of the binary tree.
LeetCode 2415 Solutions in Java, C++, Python
class Solution {
public TreeNode reverseOddLevels(TreeNode root) {
reverseOddLevelsHelper(root, 1);
return root;
}
private void reverseOddLevelsHelper(TreeNode node, int level) {
if (node == null) {
return;
}
if (level % 2 == 1) {
reverseValues(node);
}
reverseOddLevelsHelper(node.left, level + 1);
reverseOddLevelsHelper(node.right, level + 1);
}
private void reverseValues(TreeNode node) {
if (node == null) {
return;
}
TreeNode temp = node.left;
node.left = node.right;
node.right = temp;
reverseValues(node.left);
reverseValues(node.right);
}
}
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