# Easy to understand JAVA solution

• ``````public class Solution {
public int kthSmallest(TreeNode root, int k) {
int leftNum = countNodes(root.left);

if (leftNum + 1 == k)
return root.val;
else if (leftNum  + 1 > k) {
return kthSmallest(root.left, k);
}
else
return kthSmallest(root.right, k - leftNum - 1);
}

private int countNodes(TreeNode root) {
if (root == null)
return 0;

return 1 + countNodes(root.left) + countNodes(root.right);
}
}
``````

The idea is to determine where the target falls (left side, the root, or the right side) at each level.

• Nice solution! What is the worst case time complexity of your method? Still O(n)?

• This solution is much better than taking the whole in-order traversal. Asymptotically, if the tree is balanced, since countNodes is doing n/2 work, time complexity is still O(n) and not O(height)

• I think the time complexity will be huge for this solution. Count Nodes get called many times, right?

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