take two stacks of O(logn) space use one to traverse as left - > root->
right (inorder) and other as right->root-> left(reverse inorder) like
while (true){
a=top(s1);
b=top(s2);
if(a+b==sum ) break;
if(a+b <sum){
pop(s1);
}else if(a+b > sum) pop(s2);
}
if(s1 && s2 are empty) no solution
else result = top(s1) and top(s2);
On Tue, Mar 5, 2013 at 1:24 PM, marti <[email protected]> wrote:
> Given a value , print two nodes that sum to the value in a BST and normal
> tree.. time:O(n), space O(logn).
>
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Regards,
Sourabh Kumar Jain
+91-8971547841
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