Is it possible to compare two binary trees in less than O(n log n) time?
I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time.
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if ( p == null && q==null)
return true;
if (p == null || q == null)
return false;
if ( (p.val == q.val) && isSameTree(p.left, q.left) &&
isSameTree(p.right, q.right))
return true;
else
return false;
}
}
My code takes O(n log n) time.
How to approach reducing the time required?
java algorithm time-complexity binary-tree
edited 4 hours ago
nullpointer
43.8k1093179
asked 5 hours ago
Louise
411
add a comment |
I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time.
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if ( p == null && q==null)
return true;
if (p == null || q == null)
return false;
if ( (p.val == q.val) && isSameTree(p.left, q.left) &&
isSameTree(p.right, q.right))
return true;
else
return false;
}
}
My code takes O(n log n) time.
How to approach reducing the time required?
java algorithm time-complexity binary-tree
edited 4 hours ago
nullpointer
43.8k1093179
asked 5 hours ago
Louise
411
If you happen to have asizevariable at the base of the structure, compare that first.
– Boann
2 hours ago
add a comment |
I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time.
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if ( p == null && q==null)
return true;
if (p == null || q == null)
return false;
if ( (p.val == q.val) && isSameTree(p.left, q.left) &&
isSameTree(p.right, q.right))
return true;
else
return false;
}
}
My code takes O(n log n) time.
How to approach reducing the time required?
java algorithm time-complexity binary-tree
edited 4 hours ago
nullpointer
43.8k1093179
asked 5 hours ago
Louise
411
I wrote a java routine to compare 2 binary trees. I am looking for better algorithms that run in less time.
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if ( p == null && q==null)
return true;
if (p == null || q == null)
return false;
if ( (p.val == q.val) && isSameTree(p.left, q.left) &&
isSameTree(p.right, q.right))
return true;
else
return false;
}
}
My code takes O(n log n) time.
How to approach reducing the time required?
java algorithm time-complexity binary-tree
java algorithm time-complexity binary-tree
edited 4 hours ago
nullpointer
43.8k1093179
asked 5 hours ago
Louise
411
edited 4 hours ago
nullpointer
43.8k1093179
asked 5 hours ago
Louise
411
edited 4 hours ago
nullpointer
43.8k1093179
edited 4 hours ago
nullpointer
43.8k1093179
edited 4 hours ago
nullpointer
43.8k1093179
43.8k1093179
asked 5 hours ago
Louise
411
asked 5 hours ago
Louise
411
asked 5 hours ago
Louise
411
411
If you happen to have asizevariable at the base of the structure, compare that first.
– Boann
2 hours ago
add a comment |
If you happen to have asizevariable at the base of the structure, compare that first.
– Boann
2 hours ago
If you happen to have a
size variable at the base of the structure, compare that first.– Boann
2 hours ago
If you happen to have a
size variable at the base of the structure, compare that first.– Boann
2 hours ago
add a comment |
1 Answer
1
active
oldest
votes
The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.
Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).
Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.
edited 3 hours ago
ruakh
124k13197251
answered 4 hours ago
nullpointer
43.8k1093179
3
In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.
Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).
Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.
edited 3 hours ago
ruakh
124k13197251
answered 4 hours ago
nullpointer
43.8k1093179
3
In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago
add a comment |
The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.
Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).
Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.
edited 3 hours ago
ruakh
124k13197251
answered 4 hours ago
nullpointer
43.8k1093179
3
In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago
add a comment |
The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.
Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).
Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.
edited 3 hours ago
ruakh
124k13197251
answered 4 hours ago
nullpointer
43.8k1093179
The current runtime of your approach is actually O(n), where n should be the number of nodes of the tree with lesser(or if they're equal) nodes.
Also, note to compare all the values of a data structure you would have to visit all of them and that is the runtime you could achieve and not reduce further. In the current case, at the worst, you would have to visit all the nodes of the smaller tree and hence O(n).
Hence any other approach though might help you with conditional optimization, your current solution has an optimal runtime which cannot be reduced further.
edited 3 hours ago
ruakh
124k13197251
answered 4 hours ago
nullpointer
43.8k1093179
edited 3 hours ago
ruakh
124k13197251
edited 3 hours ago
ruakh
124k13197251
edited 3 hours ago
ruakh
124k13197251
124k13197251
answered 4 hours ago
nullpointer
43.8k1093179
answered 4 hours ago
nullpointer
43.8k1093179
answered 4 hours ago
nullpointer
43.8k1093179
43.8k1093179
3
In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago
add a comment |
3
In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago
3
3
In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago
In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
4 hours ago
add a comment |
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If you happen to have a
sizevariable at the base of the structure, compare that first.– Boann
2 hours ago