# Write The Shortest Program To Check If A Binary Tree Is Balanced

For each node in a balanced binary tree, the maximum difference in the heights of the left child subtree and the right child subtree are at most 1.

The height of a binary tree is the distance from the root node to the node child that is farthest from the root.

Below is an example:

           2 <-- root: Height 1
/ \
7   5 <-- Height 2
/ \   \
2   6   9 <-- Height 3
/ \  /
5  11 4 <-- Height 4


Height of binary tree: 4

The following are binary trees and a report on whether or not they are balanced:

The tree above is unbalanced.

The above tree is balanced.

Write the shortest program possible that accepts as input the root of a binary tree and returns a falsey value if the tree is unbalanced and a truthy value if the tree is balanced.

Input

The root of a binary tree. This may be in the form of a reference to the root object or even a list that is a valid representation of a binary tree.

Output

Returns truthy value: If the tree is balanced

Returns falsey value: If the tree is unbalanced.

Definition of a Binary Tree

A tree is an object that contains a value and either two other trees or pointers to them.

The structure of the binary tree looks something like the following:

typedef struct T
{
struct T *l;
struct T *r;
int v;
}T;


If using a list representation for a binary tree, it may look something like the following:

[root_value, left_node, right_node]

• May input be empty tree? – tsh Aug 6 at 2:43
• In your initial example of a tree, if you remove the leaf 4, is the remaining tree balanced? – Neil Aug 6 at 8:43
• No, not that example, I meant the initial one, using ASCII art. – Neil Aug 6 at 23:55
• According to my own implementation "C, 117 bytes": No, since the height of the right subarm tree starting from "5" is 2 and the height of the left subarm tree is 0. – T. Salim Aug 7 at 0:00
• Edits are at least 6 chars but please remove the comma from between 'balanced' and 'binary' - 'binary tree' is a noun phrase, so writing 'balanced, binary tree' is the equivalent of 'red, snow mobile' - the comma is not required. – Geza Kerecsenyi Aug 7 at 14:59

# Jelly, 11 bytes

ḊµŒḊ€IỊ;ß€Ạ


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The empty tree is represented by [].

• Thanks Erik for being amongst the first to answer this question. Jelly certainly is a very popular language on this site. I think I should take the liberty to implement this language. Good to learn from a robust golf-scripting language. – T. Salim Aug 5 at 19:14
• Congrats Erik the Outgolfer, you are winner. – T. Salim Aug 9 at 1:29

# Prolog (SWI), 49 bytes

N+_/B/C:-X+B,Y+C,abs(X-Y)<2,N is max(X,Y)+1.
0+e.


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Represents trees as Value/Left_Child/Right_Child, with the empty tree being the atom e. Defines +/2, which outputs through success or failure, with an unbound variable (or one already equal to the tree's height) on the left and the tree on the right--if the height argument is unacceptable, add 9 bytes to define -T:-_+T..

N + _/B/C :-            % If the second argument is a tree of the form _Value/B/C,
X+B,                % X is the height of its left child which is balanced,
Y+C,                % Y is the height of its right child which is balanced,
abs(X-Y) < 2,       % the absolute difference between X and Y is strictly less than 2,
N is max(X,Y)+1.    % and N is the height of the full tree.
0 + e.                  % If, on the other hand, the second argument is e, the first is 0.

• (If the value of each node could be omitted from the input, _/ could be taken out for -2 bytes.) – Unrelated String Aug 6 at 3:50

# Wolfram Language (Mathematica), 50 bytes

f@_[x_,y_]:=f@x&&f@y&&-2<Depth@x-Depth@y<2;f@_=1>0


Use Null for null, value[left, right] for nodes. For example, the following tree is written as 2[7[2[Null, Null], 6[5[Null, Null], 11[Null, Null]]], 5[Null, 9[4[Null, Null], Null]]].

    2
/ \
7   5
/ \   \
2   6   9
/ \  /
5  11 4


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• That's really pretty! – Greg Martin Aug 6 at 8:43

# Python 3.8 (pre-release), 133 125 bytes

b=lambda t:((max(l[0],r[0])+1,abs(l[0]-r[0])<2)if(l:=b(t[1]))[1]and(r:=b(t[2]))[1]else(0,0))if t else(0,1)
h=lambda t:b(t)[1]


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Takes a tree in the "list" format: A node is [value, left, right] with left and right being nodes.

Invoke the function h.

Returns 0 or False for an unbalanced tree. Returns 1 or True for a balanced tree.

## Ungolfed:

# Returns tuple (current height, subtrees are balanced (or not))
def balanced(tree):
if tree: # [] evaluates to False
left = balanced(tree[1])
right = balanced(tree[2])
# If  the subtrees are not both balanced, nothing to do, just pass it up
if left[1] and right[1]:
height = max(left[0], right[0]) + 1
subtrees_balanced = abs(left[0] - right[0]) < 2
else:
height = 0 # Value doesn't matter, will be ignored
subtrees_balanced = False
else:
height = 0
subtrees_balanced = True
return (height, subtrees_balanced)

def h(tree):
return balanced(tree)[1]


-10: Reversed logic to get rid of nots

If taking arguments in the middle of a call is allowed, this could be shortened to (115 bytes)

(b:=lambda t:((max(l[0],r[0])+1,abs(l[0]-r[0])<2)if(l:=b(t[1]))[1]and(r:=b(t[2]))[1]else(0,0))if t else(0,1))(_)[1]


with _ being the tree to check.

# JavaScript (Node.js), 49 bytes

h=([l,r])=>l?(d=h(l)-h(r))*d<2?1+h(d>0?l:r):NaN:1


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-9 bytes by Arnauld.

# JavaScript, 58 bytes

h=([l,r])=>l?(l=h(l),r=h(r),m=l>r?l:r,m+m-l-r<2?m+1:NaN):1


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Use [] for null, and [left, right, value] for nodes.

# JavaScript, 162 bytes

f=x=>{for(f=0,s=[[x,1]];s[0];){if(!((d=(t=s.pop())[0]).a&&d.b||f))f=t[1];if(f&&t[1]-f>1)return 0;if(d.a)s.push([d.a,t[1]+1]);if(d.b)s.push([d.b,t[1]+1])}return 1}


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The format of the input is an object

root={a:{node},b:{node},c:value}


## Explanation

for(f=0,s=[[x,1]];s[0];){if(!((d=(t=s.pop())[0]).a&&d.b||f))f=t[1]


Performing breadth first search find the depth of the first node which is missing one or more branches.

if(f&&t[1]-f>1)return 0;if(d.a)s.push([d.a,t[1]+1]);if(d.b)s.push([d.b,t[1]+1])}


Continuing the breadth first search, return zero if any element is two deeper than the depth of the first node missing branches.

return 1}


If no such node is found, return 1

• There is probably some way to do the breadth first search better but I couldn't think of it. – fəˈnɛtɪk Aug 5 at 20:01
• I think this fails for some valid cases such as the very first example which should become balanced when you remove the leaf 4. – Neil Aug 6 at 8:56

# Julia, 56 bytes

f(t)=t!=()&&(-(f.(t.c)...)^2<2 ? maximum(f,t.c)+1 : NaN)


With the following struct representing the binary tree:

struct Tree
c::NTuple{2,Union{Tree,Tuple{}}}
v::Int
end


c is a tuple representing the left and right nodes and the empty tuple () is used to signal the absence of a node.

Falsey value is NaN, any integer is truthy.

• Assuming the encoding is UTF-8, this is actually 57 bytes because of the ≢, according to TIO's built-in byte counter. Anyhow, welcome to CG&CC! – Unrelated String Aug 6 at 19:42
• Yes, you're right. I corrected it, so that it's now actually 56 bytes – user3263164 Aug 6 at 19:49

# Kotlin, 67 bytes

fun N.c():Int=maxOf(l?.c()?:0,r?.c()?:0)+1
fun N.b()=l?.c()==r?.c()


Where

data class N(val l: N? = null, val r: N? = null, val v: Int = 0)


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# C, 117 bytes

h(T*r){r=r?1+h(h(r->l)>h(r->r)?r->l:r->r):0;}b(T*r){return r->l&&!b(r->l)||r->r&&!b(r->r)?0:abs(h(r->l)-h(r->r))<=1;}


Struct implementation is the following:

 typedef struct T
{
struct T * l;

struct T * r;

int v;

} T;


Try This on JDoodle

• This appears to be 117 bytes, though you can do <2 for that last check instead – Jo King Aug 6 at 22:36
• Also, I'm not sure how valid this is, since it relies on a data structure defined outside of the submission – Jo King Aug 7 at 0:06

# Python 2, 9996 94 bytes

lambda A:A==[]or(abs(D(A[1])-D(A[2]))<2)*f(A[1])*f(A[2])
D=lambda A:A>[]and-~max(map(D,A[1:]))


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3 bytes from Jo King.

Takes input as: empty node is [], and other nodes are [<value>, <leftNode>, <rightNode>]. Outputs 0/1 for False/True.