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 2 mins ago
GearonGearon
101
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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 2 mins ago
GearonGearon
101
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 2 mins ago
GearonGearon
101
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 2 mins ago
GearonGearon
101
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 2 mins ago
GearonGearon
101
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 2 mins ago
GearonGearon
101
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 2 mins ago
GearonGearon
101
asked 2 mins ago
GearonGearon
101
101
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 |
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