Check whether tree1 has complete structure of tree2
0
$begingroup$
As title described, the code is written to solve the problem.
public class Main {
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
public boolean HasSubtree(TreeNode root1,TreeNode root2) {
if(root2 == null) {
return true;
}
if(root1 == null) {
return false;
}
if(root1.val == root2.val) {
return HasSubtree(root1.left, root2.left) && HasSubtree(root1.right, root2.right);
}else {
return HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
}
}
}
The algorithm I used can't pass online judgement, where
tree
asked 4 mins ago
GearonGearon
1011
New contributor
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
$begingroup$
As title described, the code is written to solve the problem.
public class Main {
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
public boolean HasSubtree(TreeNode root1,TreeNode root2) {
if(root2 == null) {
return true;
}
if(root1 == null) {
return false;
}
if(root1.val == root2.val) {
return HasSubtree(root1.left, root2.left) && HasSubtree(root1.right, root2.right);
}else {
return HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
}
}
}
The algorithm I used can't pass online judgement, where
tree
asked 4 mins ago
GearonGearon
1011
New contributor
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
0
0
$begingroup$
As title described, the code is written to solve the problem.
public class Main {
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
public boolean HasSubtree(TreeNode root1,TreeNode root2) {
if(root2 == null) {
return true;
}
if(root1 == null) {
return false;
}
if(root1.val == root2.val) {
return HasSubtree(root1.left, root2.left) && HasSubtree(root1.right, root2.right);
}else {
return HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
}
}
}
The algorithm I used can't pass online judgement, where
tree
asked 4 mins ago
GearonGearon
1011
New contributor
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As title described, the code is written to solve the problem.
public class Main {
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
public boolean HasSubtree(TreeNode root1,TreeNode root2) {
if(root2 == null) {
return true;
}
if(root1 == null) {
return false;
}
if(root1.val == root2.val) {
return HasSubtree(root1.left, root2.left) && HasSubtree(root1.right, root2.right);
}else {
return HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
}
}
}
The algorithm I used can't pass online judgement, where
tree
tree
asked 4 mins ago
GearonGearon
1011
New contributor
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 mins ago
GearonGearon
1011
New contributor
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 mins ago
GearonGearon
1011
New contributor
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 mins ago
GearonGearon
1011
asked 4 mins ago
GearonGearon
1011
1011
New contributor
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Gearon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
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: "196"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
});
}
});
Gearon is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f212447%2fcheck-whether-tree1-has-complete-structure-of-tree2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Gearon is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Gearon is a new contributor. Be nice, and check out our Code of Conduct.
Gearon is a new contributor. Be nice, and check out our Code of Conduct.
Gearon is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Code Review Stack Exchange!
- 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.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f212447%2fcheck-whether-tree1-has-complete-structure-of-tree2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
