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authorJack Sangdahl <[email protected]>2024-10-23 21:31:55 -0600
committerJack Sangdahl <[email protected]>2024-10-23 21:31:55 -0600
commit0297643f9bc94881eb3ea3b3f91741fcf34a4d7d (patch)
tree06abdc297108330fd2b5cc1c9044c1eed352e068
parent9b1afed7980a5eb2b1f4bd886d07340629fde40c (diff)
downloadleetcode-0297643f9bc94881eb3ea3b3f91741fcf34a4d7d.tar.gz
leetcode-0297643f9bc94881eb3ea3b3f91741fcf34a4d7d.zip
951. flip equivalent binary trees
-rw-r--r--flip_equivalent_binary_tree.c94
-rw-r--r--flip_equivalent_binary_tree.py57
2 files changed, 151 insertions, 0 deletions
diff --git a/flip_equivalent_binary_tree.c b/flip_equivalent_binary_tree.c
new file mode 100644
index 0000000..fedfe5c
--- /dev/null
+++ b/flip_equivalent_binary_tree.c
@@ -0,0 +1,94 @@
+// 951. Flip Equivalent Binary Trees
+#include "leetcode.h"
+
+/*
+ * for a binarytree T, we can define a flip operation as follows: choose any
+ * node, and swap the left and right child subtrees. a binary tree X is flip
+ * equivalent to a binary tree Y if and only if we can make X equal to Y after
+ * some number of flip operations. given the roots of two binary trees 'root1'
+ * and 'root2', return 'true' if the two trees are flip equivalent.
+ */
+
+//__attribute((naked))
+bool flipEquiv(struct TreeNode *root1, struct TreeNode *root2) {
+ /*
+ __asm("push %rbp;"
+ "mov %rsp,%rbp;"
+ "push %rbx;"
+ "push %r8;"
+ "push %r9;"
+ "xor %eax,%eax;"
+ "test %rdi,%rdi;"
+ "jnz l1;"
+ "test %rsi,%rsi;"
+ "jnz l0;"
+ "add $0x1,%eax;"
+ "jmp l0;"
+ "l1:;"
+ "test %rsi,%rsi;"
+ "jz l0;"
+ "mov (%rdi),%ecx;"
+ "cmp %ecx,(%rsi);"
+ "jne l0;"
+ "mov %rdi,%r8;"
+ "mov %rsi,%r9;"
+ "mov 0x8(%r8),%rdi;"
+ "mov 0x8(%r9),%rsi;"
+ "call flipEquiv;"
+ "xchg %eax,%ebx;"
+ "mov 0x10(%r8),%rdi;"
+ "mov 0x10(%r9),%rsi;"
+ "call flipEquiv;"
+ "and %ebx,%eax;"
+ "jnz l0;"
+ "mov 0x8(%r8),%rdi;"
+ "mov 0x10(%r9),%rsi;"
+ "call flipEquiv;"
+ "xchg %eax,%ebx;"
+ "mov 0x10(%r8),%rdi;"
+ "mov 0x8(%r9),%rsi;"
+ "call flipEquiv;"
+ "and %ebx,%eax;"
+ "l0:;"
+ "pop %r9;"
+ "pop %r8;"
+ "pop %rbx;"
+ "mov %rbp,%rsp;"
+ "pop %rbp;"
+ "pop %rcx;"
+ "jmp %rcx;");
+ */
+ if (!root1 || !root2)
+ return root1 == root2;
+ if (root1->val == root2->val) {
+ return (flipEquiv(root1->left, root2->right) &
+ flipEquiv(root1->right, root2->left)) ||
+ (flipEquiv(root1->left, root2->left) &
+ flipEquiv(root1->right, root2->right));
+ }
+ return false;
+}
+
+int main() {
+ struct TreeNode *r1 = treenode_create(1);
+ r1->left = treenode_create(2);
+ r1->right = treenode_create(3);
+ r1->left->left = treenode_create(4);
+ r1->left->right = treenode_create(5);
+ r1->right->left = treenode_create(6);
+ r1->right->right = NULL;
+ r1->left->right->left = treenode_create(7);
+ r1->left->right->right = treenode_create(8);
+ struct TreeNode *r2 = treenode_create(1);
+ r2->left = treenode_create(3);
+ r2->right = treenode_create(2);
+ r2->left->left = NULL;
+ r2->left->right = treenode_create(6);
+ r2->right->left = treenode_create(4);
+ r2->right->right = treenode_create(5);
+ r2->right->right->left = treenode_create(8);
+ r2->right->right->right = treenode_create(7);
+ printf("%d\n", flipEquiv(r1, r2)); // expect: 1
+ treenode_free(r1);
+ treenode_free(r2);
+}
diff --git a/flip_equivalent_binary_tree.py b/flip_equivalent_binary_tree.py
new file mode 100644
index 0000000..0d80f62
--- /dev/null
+++ b/flip_equivalent_binary_tree.py
@@ -0,0 +1,57 @@
+# 951. Flip Equivalent Binary Trees
+
+"""
+for a binarytree T, we can define a flip operation as follows: choose any
+node, and swap the left and right child subtrees. a binary tree X is flip
+equivalent to a binary tree Y if and only if we can make X equal to Y after
+some number of flip operations. given the roots of two binary trees 'root1'
+and 'root2', return 'true' if the two trees are flip equivalent.
+"""
+
+
+# Definition for a binary tree node.
+class TreeNode(object):
+ def __init__(self, val=0, left=None, right=None):
+ self.val = val
+ self.left = left
+ self.right = right
+
+
+class Solution(object):
+ def flipEquiv(self, root1, root2):
+ """
+ :type root1: Optional[TreeNode]
+ :type root2: Optional[TreeNode]
+ :rtype: bool
+ """
+ s1, s2 = [root1], [root2]
+ while s1 and s2:
+ n1, n2 = s1.pop(), s2.pop()
+ if n1 is None or n2 is None:
+ continue
+ elif not n1 or not n1 or n1.val != n2.val:
+ return False
+ if (
+ n1.left is None
+ or n2.left is None
+ or n1.left
+ and n2.left
+ and n1.left.val == n2.left.val
+ ):
+ s1.extend([n1.left, n1.right])
+ else:
+ s1.extend([n1.right, n1.left])
+ s2.extend([n2.left, n2.right])
+ return not s1 and not s2
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(
+ obj.flipEquiv(
+ root1=[1, 2, 3, 4, 5, 6, None, None, None, 7, 8],
+ root2=[1, 3, 2, None, 6, 4, 5, None, None, None, None, 8, 7],
+ )
+ )
+ print(obj.flipEquiv(root1=[], root2=[]))
+ print(obj.flipEquiv(root1=[], root2=[1]))