1
00:00:01,420 --> 00:00:02,050
Hi, everyone.

2
00:00:02,080 --> 00:00:06,290
So in this video, we're going to solve this question, identical binary.

3
00:00:06,700 --> 00:00:08,650
OK, so this is the problem statement.

4
00:00:08,900 --> 00:00:14,860
So given to minorities, what we have to do, we have to check whether the two binary trees are equal

5
00:00:14,860 --> 00:00:15,340
or not.

6
00:00:15,430 --> 00:00:17,980
OK, so what is the meaning of equal to equal means?

7
00:00:17,980 --> 00:00:21,920
Their value should be C each value should be the same and destructive should be.

8
00:00:22,240 --> 00:00:27,730
OK, I will call to Binit recall when all the values are the same and the structure of the two minorities

9
00:00:27,730 --> 00:00:31,620
are saying, OK, let us consider this example so to buy.

10
00:00:31,960 --> 00:00:32,830
So this is driven.

11
00:00:33,780 --> 00:00:38,910
And this is true, too, you can see, so driven by the values are all the same and the structures also

12
00:00:38,910 --> 00:00:40,620
seem so, in this case I will return.

13
00:00:40,620 --> 00:00:40,950
True.

14
00:00:41,040 --> 00:00:43,070
OK, so that will be billion.

15
00:00:43,800 --> 00:00:45,330
I have to return the value.

16
00:00:45,330 --> 00:00:49,630
Whether there will buy new trees, I recall, or to buy any trees are not equal.

17
00:00:49,680 --> 00:00:49,960
OK.

18
00:00:49,980 --> 00:00:50,700
True or false.

19
00:00:51,740 --> 00:00:53,130
Let us consider this example.

20
00:00:53,360 --> 00:00:57,150
So basically the same one went to two, but the structure is different, OK?

21
00:00:57,170 --> 00:01:00,620
So we're here to light in the laboratory and here to it.

22
00:01:00,650 --> 00:01:01,340
Right subtree.

23
00:01:01,550 --> 00:01:03,250
So in this case, I will return false.

24
00:01:03,440 --> 00:01:05,510
They are not identical by nuclear.

25
00:01:05,680 --> 00:01:11,320
OK, let us consider this example so they doctor same structure structuralism.

26
00:01:11,810 --> 00:01:19,220
Now one one they are saying, OK, this no see here to the left hand side, but here too is present

27
00:01:19,220 --> 00:01:20,140
in the right hand side.

28
00:01:20,180 --> 00:01:21,290
So the values are different.

29
00:01:21,470 --> 00:01:26,600
OK, all of the structure, same structure, same this one root node then left and right.

30
00:01:26,600 --> 00:01:29,240
And similarly, this also has the same structure.

31
00:01:29,310 --> 00:01:30,640
OK, same.

32
00:01:30,680 --> 00:01:32,950
But the values are different here to this present here.

33
00:01:32,960 --> 00:01:36,130
One is Besant similarly in the right to play here when it's present.

34
00:01:36,140 --> 00:01:37,290
But here who is present.

35
00:01:37,520 --> 00:01:39,380
So in this case, I will return false.

36
00:01:40,690 --> 00:01:43,690
OK, so the question is very simple, given to minorities.

37
00:01:43,720 --> 00:01:47,260
I have to check, I have to return to boolean value, whether to recycle or not.

38
00:01:47,590 --> 00:01:49,180
So this is a very simple question.

39
00:01:49,240 --> 00:01:54,000
OK, so how we can solve this question so we can solve this question with the help of recursion.

40
00:01:54,070 --> 00:01:57,970
OK, we can very easily solve this question with the help of recursion.

41
00:01:58,180 --> 00:01:59,440
So consider Daughtry's.

42
00:01:59,440 --> 00:02:06,100
So this is driven the size of the plane ride subtree and similarly, I have to do so.

43
00:02:06,130 --> 00:02:07,460
It will have its left and right.

44
00:02:07,460 --> 00:02:08,020
Sudbury's.

45
00:02:08,080 --> 00:02:12,700
OK, so when these two trees will be equal, very simple condition.

46
00:02:13,240 --> 00:02:15,530
So everyone's left should be close to.

47
00:02:15,610 --> 00:02:21,210
So first is basically the left should be close to the left.

48
00:02:21,610 --> 00:02:23,470
Second is basically the ones right.

49
00:02:23,470 --> 00:02:24,850
Should be close to Titos.

50
00:02:24,850 --> 00:02:25,160
Right.

51
00:02:25,780 --> 00:02:26,710
So dittos.

52
00:02:27,990 --> 00:02:33,540
Divorce rate should be close to details, right, and the third is basically the value should be same.

53
00:02:33,570 --> 00:02:37,200
OK, so basically the one value should be close to it.

54
00:02:37,230 --> 00:02:43,050
Well, OK, I'm repeating myself when the Daughtry's will be equal.

55
00:02:44,650 --> 00:02:48,910
When the two trains will be equal there, a very simple, simple condition, their values should be

56
00:02:48,930 --> 00:02:51,550
same, the value should be equal to value.

57
00:02:52,850 --> 00:02:59,610
Even left subtree should be close to details left Saberi, Devens right, Saberi should be close to

58
00:02:59,610 --> 00:02:59,930
details.

59
00:02:59,940 --> 00:03:00,660
Right, Saberi.

60
00:03:00,680 --> 00:03:01,010
OK.

61
00:03:03,220 --> 00:03:07,590
Now, how we can check whether we are equal or not, basically we will call recursion on it.

62
00:03:08,700 --> 00:03:11,490
We will call recursion on it simple.

63
00:03:11,910 --> 00:03:14,280
OK, we can write the code.

64
00:03:14,310 --> 00:03:16,390
OK, so now let us likely write the code.

65
00:03:17,460 --> 00:03:19,380
So this is the question, same tree.

66
00:03:19,430 --> 00:03:25,200
OK, so after Rotana billion value, the name of the function is is same tree and it is sticking to

67
00:03:25,200 --> 00:03:27,450
trees as input B and given the two trees.

68
00:03:27,480 --> 00:03:31,770
OK, so let's change the name and be OK.

69
00:03:33,210 --> 00:03:34,530
And we are the Daughtry's.

70
00:03:35,500 --> 00:03:36,160
So first.

71
00:03:37,550 --> 00:03:44,030
Let us write very simple, simple condition, so we've basically driven Isdell, OK, three one does

72
00:03:44,030 --> 00:03:47,020
not exist and Preto does not exist.

73
00:03:47,180 --> 00:03:49,190
So basically reasonable.

74
00:03:50,710 --> 00:03:53,910
And tribesmen, if the tunnel, that means they are equal.

75
00:03:54,040 --> 00:03:57,100
So if they are equal, I can return to simple.

76
00:03:58,020 --> 00:04:00,330
Now, if everyone isn't Al.

77
00:04:03,140 --> 00:04:10,520
Driven, seasonal, but Priebke is not only so Truby is normal and reasonable.

78
00:04:10,550 --> 00:04:11,750
So this is very simple.

79
00:04:12,020 --> 00:04:14,830
That means that both the trees are not equal because one is none.

80
00:04:15,080 --> 00:04:16,160
Other one is normal.

81
00:04:16,670 --> 00:04:17,750
So I can return false.

82
00:04:19,120 --> 00:04:22,930
Similarly, the other way around IFPRI is Dardanelle.

83
00:04:24,950 --> 00:04:26,410
And drib is Nele.

84
00:04:27,340 --> 00:04:33,600
So if this is a scenario, then also I can return false because they will not be OK when this one is

85
00:04:33,610 --> 00:04:34,070
not null.

86
00:04:35,230 --> 00:04:41,580
Now, what is the condition that he and Drillbit they are seeing, so I will use these three conditions,

87
00:04:41,620 --> 00:04:43,870
OK, I will use these three conditions.

88
00:04:43,900 --> 00:04:46,800
So if all the conditions are true, I will return to those.

89
00:04:46,810 --> 00:04:47,650
I will return false.

90
00:04:47,680 --> 00:04:51,430
OK, so if all the conditions are true, I will return true.

91
00:04:51,430 --> 00:04:54,460
Otherwise I will return false if any of the condition is false.

92
00:04:55,270 --> 00:04:56,800
OK, so.

93
00:04:57,760 --> 00:05:01,750
Let's so what should happen so evalu.

94
00:05:03,340 --> 00:05:04,770
Should be close to be value.

95
00:05:05,910 --> 00:05:08,010
OK, their value should be simple.

96
00:05:08,870 --> 00:05:14,310
This is the first condition, second condition is basically their left subtree should also be seen.

97
00:05:14,360 --> 00:05:17,230
So what we will do, we will take the function of a cemetery.

98
00:05:17,280 --> 00:05:20,670
OK, this is a recursive function, so let's copy them.

99
00:05:20,670 --> 00:05:21,300
Function name.

100
00:05:24,870 --> 00:05:32,730
So is same tree, I will give it to trees, so I am giving the leaves a tree of tree and I'm giving

101
00:05:32,730 --> 00:05:33,940
the A of tree.

102
00:05:34,710 --> 00:05:37,970
OK, so this function is same tree function.

103
00:05:37,980 --> 00:05:42,460
This will be true or false, whether through trees e left and be left with the same or not.

104
00:05:42,480 --> 00:05:48,210
OK, this is recursive function and similarly I will call this function the same tree for the right

105
00:05:48,330 --> 00:05:48,780
price.

106
00:05:48,810 --> 00:05:53,880
OK, so I will, I have to give two trees so erate and be right.

107
00:05:56,380 --> 00:06:01,930
So we've basically all the three conditions are true if this condition is true, if this basically.

108
00:06:04,290 --> 00:06:09,630
So what is the work of dysfunction, so what is the work of is the same dysfunction?

109
00:06:10,050 --> 00:06:16,170
So it will take two days as input and B and it will return a boolean value, whether they're three a.m.

110
00:06:16,590 --> 00:06:18,090
or three a.m..

111
00:06:18,280 --> 00:06:21,060
OK, so these are very simple, simple conditions.

112
00:06:21,510 --> 00:06:25,080
Now, what I'm doing here is so when are the two trees will be equal?

113
00:06:25,830 --> 00:06:27,240
When two trees will be equal?

114
00:06:28,540 --> 00:06:34,420
When their values are the same, so if the values of A, B, if their values the same, if they live

115
00:06:34,420 --> 00:06:37,870
surprising, so if they are left separately the same.

116
00:06:38,080 --> 00:06:42,760
So but I will do I will call this function the same tree and I will give the left's.

117
00:06:43,420 --> 00:06:45,750
I will give that left subtree of B OK.

118
00:06:45,790 --> 00:06:51,070
It will return true or false requestion will handle OK because it will handle everything.

119
00:06:52,080 --> 00:06:58,650
Now, again, there should also be some so I'm calling the function the same tree, I'm calling the

120
00:06:58,650 --> 00:07:00,960
function the same tree, and I am passing their rates up.

121
00:07:01,230 --> 00:07:04,640
So I'm giving the right set of I am giving the right set of me.

122
00:07:04,650 --> 00:07:07,440
OK, now every I'm using and operated.

123
00:07:07,470 --> 00:07:10,620
OK, so if all the conditions are true.

124
00:07:12,040 --> 00:07:15,560
That means and B, they both have saying I can return, true.

125
00:07:15,580 --> 00:07:16,390
I can return true.

126
00:07:16,390 --> 00:07:17,740
Otherwise I will return false.

127
00:07:17,750 --> 00:07:18,310
Simple.

128
00:07:20,550 --> 00:07:21,330
So what I will do.

129
00:07:24,520 --> 00:07:29,210
I will return true in the end part, what I can do.

130
00:07:29,230 --> 00:07:30,190
I can return false.

131
00:07:31,880 --> 00:07:32,390
Simple.

132
00:07:38,610 --> 00:07:41,000
OK, our goal is working fine, Alert summered.

133
00:07:45,460 --> 00:07:47,380
OK, so our goal is working fine.

134
00:07:48,190 --> 00:07:52,090
OK, now a little bit of time, complexity and the space complexity.

135
00:07:53,040 --> 00:07:58,840
So basically three letters, let's say this containing and notes and three B is containing amnot.

136
00:07:58,870 --> 00:07:59,260
OK.

137
00:08:00,700 --> 00:08:02,060
So these are the number of nodes.

138
00:08:02,920 --> 00:08:07,330
So what I'm doing here is I'm basically going through each and every note of it and I'm going through

139
00:08:07,330 --> 00:08:09,220
each and every note of be OK.

140
00:08:09,520 --> 00:08:11,430
So all the time complexity.

141
00:08:11,440 --> 00:08:13,450
So the time complexity will be emplacing.

142
00:08:13,630 --> 00:08:16,590
OK, I'm going through each and all of that.

143
00:08:17,590 --> 00:08:23,260
But in reality, the time complexity will be basically off minimum of Immonen.

144
00:08:24,710 --> 00:08:29,240
Because if the laws are different, OK, see if the laws are different, this is a..

145
00:08:29,360 --> 00:08:30,590
This is a..

146
00:08:30,630 --> 00:08:32,330
That means the structure is different.

147
00:08:32,990 --> 00:08:35,730
OK, that means the structure is different.

148
00:08:35,960 --> 00:08:38,659
So if the structure is different with the help of these two conditions.

149
00:08:39,840 --> 00:08:43,740
With the help of these two conditions, I will return false and my reaction will stop.

150
00:08:43,770 --> 00:08:48,090
OK, so that's why I have written here the minimum of Immonen.

151
00:08:48,780 --> 00:08:52,950
We will not explore the whole lot, so I will explore these nodes.

152
00:08:52,980 --> 00:08:54,500
OK, simple.

153
00:08:54,990 --> 00:08:57,910
So if you have any doubt in this question, you can ask me, ok.

154
00:08:58,740 --> 00:08:59,860
The code is very simple.

155
00:08:59,910 --> 00:09:01,950
OK, so if you have it out you can ask me.

156
00:09:01,980 --> 00:09:02,400
Thank you.