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Program to check whether two trees can be formed by swapping nodes or not in Python
Suppose we have two trees, we have to check whether we can transform first tree into second one by swapping any node's left and right subtrees any number of times.
So, if the input is like
then the output will be True
To solve this, we will follow these steps −
que1 := a queue with root0 initially
que2 := a queue with root1 initially
while que1 and que2 are not empty, do
temp1 := a new list, temp2 := a new list
values1 := a new list, values2 := a new list
if que1 and que2 are containing different number of elements, then
return False
for i in range 0 to size of que1 − 1, do
insert value of que1[i] at the end of values1
insert value of que2[i] at the end of values2
if right of que1[i] is not null, then
insert right of que1[i] at the end of temp1
if left of que1[i] is not null, then
insert left of que1[i] at the end of temp1
if right of que2[i] is not null, then
insert right of que2[i] at the end of temp2
if left of que2[i] is not null, then
insert left of que2[i] at the end of temp2
if values1 is not same as values2, then
if values1 is not same as values2 in reverse order, then
return False
que1 := temp1, que2 := temp2
return True
Let us see the following implementation to get better understanding −
Example
class TreeNode: def __init__(self, data, left = None, right = None): self.val = data self.left = left self.right = right class Solution: def solve(self, root0, root1): que1 = [root0] que2 = [root1] while que1 and que2: temp1 = [] temp2 = [] values1 = [] values2 = [] if len(que1) != len(que2): return False for i in range(len(que1)): values1.append(que1[i].val) values2.append(que2[i].val) if que1[i].right: temp1.append(que1[i].right) if que1[i].left: temp1.append(que1[i].left) if que2[i].right: temp2.append(que2[i].right) if que2[i].left: temp2.append(que2[i].left) if values1 != values2: if values1 != values2[::-1]: return False que1 = temp1 que2 = temp2 return True ob = Solution() root = TreeNode(2) root.right = TreeNode(4) root.right.left = TreeNode(3) root.right.right = TreeNode(5) root1 = TreeNode(2) root1.left = TreeNode(4) root1.left.left = TreeNode(3) root1.left.right = TreeNode(5) print(ob.solve(root, root1))
Input
root = TreeNode(2) root.right = TreeNode(4) root.right.left = TreeNode(3) root.right.right = TreeNode(5) root1 = TreeNode(2) root1.left = TreeNode(4) root1.left.left = TreeNode(3) root1.left.right = TreeNode(5)
Output
True
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