Binary tree
Link the node pointer "s" to the next node to right side on same level, null if last node on its level.
input:
1
/ \
2 3
/ \ \
4 5 6
/ \ \
7 8 9
output:
1 -- null
/ \
2 -- 3 -- null
/ \ \
4 -- 5 --- 6 -- null
/ \ \
7 ---- 8 --- 9 --- null Explaination:
root1.s=null
root2.s=root3
root3.s=null
root4.s=root5 continue..
class NodeBT {
int val;
NodeBT left;
NodeBT right;
NodeBT s;
public NodeBT(int val) {
left=null;
right=null;
s=null;
this.val=val;
}
}