Simple Java Recursive Solution O(nlogn)

• ``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isBalanced(TreeNode root) {
if(null == root) {
return true;
}

return Math.abs(height(root.left) - height(root.right)) < 2 && isBalanced(root.left) && isBalanced(root.right);
}

public int height(TreeNode root) {
if(null == root) {
return 0;
}

return 1 + Math.max(height(root.left), height(root.right));
}
}
``````

• Can you please explain why the time complexity is O(nlogn) ? Thanks.

• My understanding is that for every node (n nodes) you do height (O(log(n))) consequently O(nlog(n))

• I don't think time complexity is right.
This is my solution .
In first level, O(1) time to split T(n) to 2T(n/2)
second level, O(2) time to split 2
T(n/2) to 4*(n/4)
Thus, time complexity is O(n).

``````public class Solution {
public boolean isBalanced(TreeNode root) {
int res = helper(root);
return res != -1;
}

private int helper(TreeNode root) {
if (root == null) {
return 0;
}
int left = helper(root.left);
int right = helper(root.right);
if (left == -1 || right == -1 || Math.abs(left - right) > 1) {
return -1;
}
return Math.max(left, right) + 1;
}
}
``````

• Complexity of Height() when called on a tree with 'n' nodes is O(n). (It visits every node once)
As we descend each level, Height() is called on subtrees with only n/2 nodes.

Lets take a complete binary tree,
When isBalanced() calls Height() on the root node -> n
When called from level 2 -> n/2 + n/2
When called from level 3 -> n/4 + n/4 + n/4 + n/4
......
When called from last level -> again n (n/L x L times)

So total complexity = n * no.of levels = O(nlogn)

PS: Just edit Height Function to remember when height difference of Child Subtrees > 1 for a O(n) solution.

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