Inverted Subtree - Google Top Interview Questions


Problem Statement :


A tree is defined to be an inversion of another tree if either:

Both trees are null
Its left and right children are optionally swapped and its left and right subtrees are inversions.
Given binary trees source and target, return whether there's some inversion T of source such that it's a subtree of target. That is, whether there's a node in target that is identically same in values and structure as T including all of its descendants.

Constraints

n ≤ 12 where n is the number of nodes in target

Example 1

Input

source = [0, [1, null, [3, null, null]], [2, null, null]]

target = [5, [2, null, null], [0, [2, null, null], [1, [3, null, null], null]]]

Output

True


Solution :



title-img 




                        Solution in C++ :

bool recur(Tree* source, Tree* target) {
    if (!source && !target) return true;
    if (!source || !target || source->val != target->val) return false;
    return ((recur(source->left, target->left) && recur(source->right, target->right)) ||
            (recur(source->left, target->right) && recur(source->right, target->left)));
}
bool solve(Tree* source, Tree* target) {
    if (!source && !target) return true;
    if (!source || !target) return false;
    if (source->val == target->val) {
        return (recur(source, target) || solve(source, target->left) ||
                solve(source, target->right));
    } else {
        return (solve(source, target->left) || solve(source, target->right));
    }
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public boolean solve(Tree source, Tree target) {
        if (target == null)
            return source == target;
        return isInv(source, target) || solve(source, target.left) || solve(source, target.right);
    }

    public boolean isInv(Tree s, Tree t) {
        if (s == null || t == null)
            return s == t;
        if (s.val != t.val)
            return false;
        return isInv(s.left, t.right) && isInv(s.right, t.left)
            || isInv(s.left, t.left) && isInv(s.right, t.right);
    }
}
                    


                        Solution in Python : 
                            
# class Tree:
#     def __init__(self, val, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def solve(self, source, target):
        if not source or not target:
            return not source and not target
        t = self.isSubtree(target, source)
        return t

    def isSubtree(self, s, t):
        def sametree(s, t):
            if not s or not t:
                return not s and not t
            elif s.val == t.val:
                return (sametree(s.left, t.left) and sametree(s.right, t.right)) or (
                    sametree(s.right, t.left) and sametree(s.left, t.right)
                )
            else:
                return False

        if not s:
            return False
        elif sametree(s, t):
            return True
        else:
            return self.isSubtree(s.right, t) or self.isSubtree(s.left, t)


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