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T1 and T2 are two very large binary trees, with T1 much bigger than T2. Create an algorithm to determine if T2 is a subtree of T1.
A tree T2 is a subtree of T1 if there exists a node n in T1 such that the subtree of n is identical to T2. That is, if you cut off the tree at node n, the two trees would be identical.
Example1:
Input: t1 = [1, 2, 3], t2 = [2] Output: true
Example2:
Input: t1 = [1, null, 2, 4], t2 = [3, 2] Output: false
Note:
Find the t2 node in t1 first, then use the depth-first search (DFS) algorithm to make sure that the subtree and the subtree of t2 are identical, otherwise return FALSE.
class Solution:
def checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:
if t1 == None:
return False
if t2 == None:
return True
return self.dfs(t1,t2) or self.checkSubTree(t1.left,t2) or self.checkSubTree(t1.right,t2)
def dfs(self, t1: TreeNode, t2: TreeNode) -> bool:
if not t1 and t2 :
return False
if not t2 and not t1:
return True
if t1.val != t2.val:
return False
else:
return self.dfs(t1.left,t2.left) and self.dfs(t1.right,t2.right)
class Solution {
public boolean checkSubTree(TreeNode t1, TreeNode t2) {
if (t2 == null)
return true;
if (t1 == null)
return false;
return isSubTree(t1, t2) || checkSubTree(t1.left, t2) || checkSubTree(t1.right, t2);
}
public boolean isSubTree(TreeNode t1, TreeNode t2){
if (t2 == null)
return true;
if (t1 == null)
return false;
if (t1.val != t2.val)
return false;
return isSubTree(t1.left,t2.left) && isSubTree(t1.right,t2.right);
}
}
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