Is it possible to compare two binary trees in less than O(n log n) time?












6














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







share|improve this question
























  • If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    2 hours ago
















6














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







share|improve this question
























  • If you happen to have a size variable at the base of the structure, compare that first.
    – Boann
    2 hours ago














6












6








6


1





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







share|improve this question















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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago









nullpointer

43.8k1093179




43.8k1093179










asked 4 hours ago









Louise

311




311












  • 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
















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












1 Answer
1






active

oldest

votes


















7














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.






share|improve this answer



















  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    3 hours ago











Your Answer






StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54067216%2fis-it-possible-to-compare-two-binary-trees-in-less-than-on-log-n-time%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7














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.






share|improve this answer



















  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    3 hours ago
















7














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.






share|improve this answer



















  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    3 hours ago














7












7








7






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.






share|improve this answer














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.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 hours ago









ruakh

124k13197251




124k13197251










answered 3 hours ago









nullpointer

43.8k1093179




43.8k1093179








  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    3 hours ago














  • 2




    In fact ... it is O(n) worst case. The best case is O(1).
    – Stephen C
    3 hours ago








2




2




In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
3 hours ago




In fact ... it is O(n) worst case. The best case is O(1).
– Stephen C
3 hours ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Stack Overflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54067216%2fis-it-possible-to-compare-two-binary-trees-in-less-than-on-log-n-time%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Eastern Orthodox Church

Image
Not to be confused with Oriental Orthodoxy, Eastern Christianity, or Eastern Catholic Churches. "Eastern Orthodoxy", "Orthodox Church", "Orthodox Catholic Church", and "Orthodox Christian Church" redirect here. For other uses, see Orthodox (disambiguation). This article may be too long to read and navigate comfortably . The readable prose size is 132 kilobytes. Please consider splitting content into sub-articles, condensing it, or adding or removing subheadings. ( January 2018 ) Part of a series on the Eastern Orthodox Church Mosaic of Christ Pantocrator, Hagia Sophia Overview Structure Theology (History of theology) Liturgy Church history Holy Mysteries View of salvation View of Mary View of icons Background Crucifixion / Resurrection / Ascension of Jesus Christianity Christian Church Apostolic succession Four Marks of the Church Orthodoxy Organization Autocephaly Patriarchate ...
Read more

Zagreb

Image
This article is about the Croatian capital city. For other uses, see Zagreb (disambiguation). City in City of Zagreb, Croatia Zagreb City Grad Zagreb City of Zagreb Clockwise, from top: St. Mark's Square, Croatian State Archives, Zagreb Tram, Art Pavilion, Cibona & HOTO towers and Croatian National Theatre. Flag Coat of arms Zagreb Location of Zagreb in Croatia Show map of Croatia Zagreb Zagreb (Europe) Show map of Europe Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Coordinates: 45°49′N 15°59′E  /  45.817°N 15.983°E  / 45.817; 15.983 Country   Croatia County City of Zagreb RC diocese 1094 Free royal city 1242 Unified 1850 Subdivisions 17 city districts 218 local committees (70 settlements) Government  • Type Mayor-Council  • Mayor Milan Bandić (BM 365)  • City Assembly 51 members BM 365, ZL, NS-R (14) SDP, GLAS, HNS, HSS, NH-PS (13) ...
Read more

Understanding the information contained in the Deep Space Network XML data?

Image
up vote 1 down vote favorite NASA's Deep Space Network activity can be viewed conveniently at https://eyes.nasa.gov/dsn/dsn.html but if you would like data in a more raw XML form, it can be accessed as described in this answer Below is a small Python script that reads the XML into a list of dictionaries, and saves it to disk as a JSON. The dictionaries are organized at the higher level by dish, then by spacecraft per dish. Question: What are the most-likely meanings of all the data field contained in the Deep Space Network XML data? This is the example url from the other answer, used in the script below: https://eyes.nasa.gov/dsn/data/1365107113.xml def dictify(r,root=True): # https://stackoverflow.com/questions/2148119/how-to-convert-an-xml-string-to-a-dictionary-in-python/30923963#30923963 if root: ret...
Read more