This is a note when I try to implement RedBlack Tree (RBT) by reverseengineering the RBT visualization at https://www.cs.usfca.edu/~galles/visualization/RedBlack.html.
What is RBT?
RBT is a binary tree that rebalancing itself. Each node in tree have one attribute, its either RED or BLACK.
Why do we need a balanced tree?
To minimize time and/or operation when doing insert, delete, and search in tree.
For example, if we insert five nodes a
, b
, c
, d
, and e
sequentially into normal unbalanced tree, we will get,
a \ b \ c \ d \ e
If we try to search for e
, we will need five comparisons, start from the top a
until e
. So, in other words it will take O(n)
.
If we use RBT, we will get a tree like these one,
b / \ a d / \ c e
and when searching for e
we only do three comparisons, b
, d
, and e
itself; in other words it will take O(log n)
.
Rules
RBT have three general rules,
1. Root must be BLACK.
2. If node is RED, both of its childs must be BLACK.
3. The number of black node from root to leaf, in every path, must be equal.
Data structure
A simple data structure for RBT in C would be like,
enum rbt_color { RED = 1 , BLACK = 2 }; struct TreeNode { rbt_color color; TreeNode* left; TreeNode* right; TreeNode* parent; void* content; }; struct RBT { TreeNode* root; };
Operations
There are three common operations for RBT: insertion, deletion, and search. Search was conducted like normal binary tree, while insertion and deletion require complex operation to make rule 3 is not violated.
Tree Rotations
Insertion and deletion will require two functions: rotateleft and rotateright.
Tree Rotate Left
Given x
as node to be rotated, rotateleft will,

make the
x
.right become parent ofx
, 
x
become left child of new parent, and 
left node of right (
x.right.left
orgleft
) become right node ofx
, replacing thex.right
node itself.
As in illustration,
... ...   x right / \ ==> / ... right x / / \ gleft ... gleft
The code for this operation would be like,
void tree_rotate_left(TreeNode* root, TreeNode* x) { TreeNode* parent = x>parent; TreeNode* right = x>right; TreeNode* gleft = right>left; if (gleft) { gleft>parent = x; } right>parent = parent; if (!parent) { root = right; } else { if (x>is_left_of(parent)) { parent>left = right; } else { parent>right = right; } } right>left = x; x>parent = right; return root; }
Tree Rotate Right
The rotateright operation is similar with rotateleft, we just swap the
left
and right
attribute with right
and left
.
As an illustration,
... ...   x left / \ \ left ... ==> x \ / \ gright gright ...
The code for rotateright operation would be like,
void tree_rotate_right(TreeNode* root, TreeNode* x) { TreeNode* parent = x>parent; TreeNode* left = x>left; TreeNode* gright = left>right; if (gright) { gright>parent = x; } left>parent = parent; if (!parent) { root = left; } else { if (x>is_left_of(parent)) { parent>left = left; } else { parent>right = left; } } left>right = x; x>parent = left; return root; }
Insertion
Inserting a new node to RBT is just like normal binary tree, except for three things,

the color of new node always RED.

If new node has the same content with node in the tree, there are two condition that we could do (depends on programmer needs): replace them or go to the right node looking for empty leaf.

After node has been inserted, the whole tree color need to be fixed up.
Since new node will be at the bottom of the tree, fixing up the tree will be start from bottom until root, so its a loop condition. The loop will stop when these two conditions met,

when parent is black, or

node does not have grandparent.
Based on these, we can start by creating insertfixup function with loop,
static TreeNode* INSERT_FIXUP(TreeNode* root, TreeNode* node) { TreeNode* parent = NULL; TreeNode* gp = NULL; // grand parent; TreeNode* auth = NULL; // this is the sibling of node's parent. parent = node>get_parent(); while (parent && parent>is_red()) { gp = node>get_grand_parent(); if (!gp) { break; } } }
Our first task is to check the color of sibling of our parent, for simplicity
lets call it as aunt
.
Lets assume that aunt is RED, with the following tree looks like,
...  gp:B / \ parent:R aunt:R / new:R
Since the tree broke the second rule, we fix it by making the grandparent RED and parent and aunt as BLACK, so the tree would be like,
...  gp:R / \ parent:B aunt:B / new:R
Continuing from our previous loop, the code for this condition would be like these,
static TreeNode* INSERT_FIXUP(TreeNode* root, TreeNode* node) { TreeNode* parent = NULL; TreeNode* gp = NULL; // grand parent; TreeNode* auth = NULL; // this is the sibling of node's parent. parent = node>get_parent(); while (parent && parent>is_red()) { gp = node>get_grand_parent(); if (!gp) { break; } if (parent>is_left_of(gp)) { aunt = gp>right; if (aunt && aunt>is_red()) { parent>set_black(); auth>set_black(); gp>set_red(); node = gp; } } else { ... } } }
What if aunt have childs? Would that make the three unbalanced? The answer is the RED aunt will never have childs. Remember, when we insert node to the tree, the node is at the bottom, which means its parent and/or aunt is at the bottom too. Even if the aunt is BLACK, the tree would be unbalance from the start because parent is RED, unless they have BLACK childs.
Our next condition is if aunt is BLACK and new node is the right child or parent, with the following illustration,
...  gp:B / \ parent:R aunt:B \ new:R
To fix the tree we rotate the parent to the left, set node to start again from bottom (node = parent), make the node’s parent into BLACK, make the grandparent into RED, and then rotate the grandparent to the right.
... ...  =>  gp:R new:B / \ / \ new:B aunt:B parent:R gp:R / \ parent:R aunt:B
The code for this condition would be like,
static TreeNode* INSERT_FIXUP(TreeNode* root, TreeNode* node) { TreeNode* parent = NULL; TreeNode* gp = NULL; // grand parent; TreeNode* auth = NULL; // this is the sibling of node's parent. parent = node>get_parent(); while (parent && parent>is_red()) { gp = node>get_grand_parent(); if (!gp) { break; } if (parent>is_left_of(gp)) { aunt = gp>right; if (aunt && aunt>is_red()) { parent>set_black(); auth>set_black(); gp>set_red(); node = gp; } else { if (node>is_right_of(parent)) { RBT.root = tree_rotate_left(RBT.root, parent); node = parent; } node>parent>set_black(); gp>set_red(); RBT.root = tree_rotate_right(RBT.root, gp); } } else { ... } } }
That is the gist for insertion.
The rest of it (where we mark with …
in the code) is just mirroring the
above code by swaping the left and right with right and left.
Deletion
Node deletion on tree have three conditions: node does not have childs, node have one child (either left or right), and node have both childs. The idea when removing node from treee was by search their replacement at the bottom, swap their content, fixing up, and then delete the node or replacement.
Node have both child
In this condition we will search the successor by finding the largest (the inner right or the rightedge) node of the left child. For example, in this three illustration,
(1) (2) (3) X X X / \ / \ / \ A ... S ... S ... / \ / \ A U A U / / \ T ... V
The successor for X in tree 1 would be A.
The successor for X in tree 2 would be U
; and the successor for X in tree 3
is V
.
In tree 2, we see that the replacement U
still have child, so we need a
recursion when removing node that have both child.
The code would be like these,
TreeNode* remove_have_both_childs(TreeNode* x) { TreeNode* heir = x>left; while (heir>right) { heir = heir>right; } heir>swap_content(x); return remove(heir); }
Node have one child
The next condition is when we want to remove a node that have child, either its left or right child. We just need to swap the left or right with the node itself and do a remove operation on the child.
TreeNode* remove(TreeNode* x) { TreeNode* left = x>left; TreeNode* right = x>right; if (left && right) { return remove_have_both_childs(x); } if (left) { left>swap_content(x); remove_with_no_child(left); return left; } if (right) { right>swap_content(x); remove_with_no_child(right); return right; } remove_with_no_child(x); return x; }
Node have no childs
There are two condition that we will meet when removing node without child. One is if child does not have parent, means its the root; second is if its color is RED. Both of this condition does not to be fixing up, just set the root to be NULL or remove the node directly.
TreeNode* remove_with_no_child(TreeNode *x) { TreeNode* parent = x>parent; if (!parent) { RBT.root = NULL; x>detach(); return x; } if (x>is_black()) { do_rebalance(x); } if (x>is_left_of(parent)) { parent>left = NULL; } else { parent>right = NULL; } x>detach(); return x; }
Finally, we came to the last algorithm on rebalancing tree for deletion.
Rebalancing Tree for Deletion
Rebalancing tree for deletion is like insertion, we do it with loop, from bottom to the the top but different stop conditions. One, stop only if node is root; second, stop when node does not have sibling.
We also have several conditions inside the loop that will break the loop immediately.
The idea for removing node is by looking into their sibling color or if sibling have childs, the color of its childs.
In this explanation we will assume that node that we will remove is always at the left of its parent. The operation for condition if node to be removed is from right of its parent is identical, just by mirroring the leftside operation.
Case 1: sibling is red. In this case we rotate the parent to the left, which make the sibling become our grandparent; and set sibling color to BLACK. If parent, after rotation, have right child then set its color to RED. After this operation we can break the loop immediately.
... => ...   p:B s:B / \ / \ x:B s:R p:B sr:B / \ / \ sl:B sr:B x:B sl:R
Case 2: sibling right child is RED. In this case we rotate the parent to the left, which make the sibling become our grandparent. If both sibling childs is RED, the sibling color become RED. Both of sibling childs’ color will become BLACK. After this operation we can break the loop immediately.
Illustration one, when sibling right child are RED and parent is BLACK,
... => ...   p:B s:B / \ / \ x:B s:B p:B sr:B / \ / \ / \ sl:R sr:R x:B sl:R ... ... / \ ... ...
Illustration two, when sibling right child are RED and parent is also RED,
... => ...   p:B s:R / \ / \ x:B s:B p:B sr:B / \ / \ sl:B sr:R x:B sl:B
Case 3: sibling left child is RED.
In this case we rotate the sibling to the right and then swap the attribute of
left child with sibling itself.
After this, we continue the operation from the x
again, which make the loop
back to Case 2.
... => ...   p:B p:B / \ / \ x:B s:B x:B sl:B / \ / \ sl:R sr:B Y s:R / \ / \ Y Z Z sr:B
Case 4: sibling have no childs or both are BLACK. In this case we set sibling color to RED. If parent is RED, set parent color to BLACK and break the loop immediately. If parent is root, break the loop immediately.
Now that we know all the cases for deletion, we can implement it to the code,
void RBT:do_rebalance(TreeNode* x) { TreeNode* parent = NULL; TreeNode* sibling = NULL; TreeNode* siblingl = NULL; TreeNode* siblingr = NULL; while (x != RBT.root) { parent = x>get_parent(); if (x>is_left_of(parent)) { sibling = parent>get_right(); if (!sibling) { break; } if (sibling>is_red()) { RBT.root = tree_rotate_left(RBT.root, parent)); sibling>set_attr_to_black(); if (parent>_right) { parent>_right>set_attr_to_red(); } return; } if (sibling>right && sibling>is_right_red()) { RBT.root = tree_rotate_left(RBT.root, parent)); if (sibling>have_red_childs()) { sibling>set_attr_to_red(); } else { sibling>set_attr_to_black(); } sibling>set_childs_attr_to_black(); return; } siblingl = sibling>get_left(); if (siblingl && siblingl>is_red()) { RBT.root = tree_rotate_right(RBT.root, sibling)); sibling>swap_attr(siblingl); parent = x; continue; } if (sibling>have_no_childs()  sibling>have_black_childs()) { sibling>set_attr_to_red(); if (parent>is_red()) { parent>set_attr_to_black(); return; } if (parent == RBT.root) { return; } } } else { ... } x = parent; } RBT.root>set_childs_attr_to_black(); }
Remember that binary tree is only have two nodes, left and right. So, any operation on the left node will identical with the right node.
If you want to see or use the full working implementation, see the libvos package in Github [1], its a open source library with BSD license.
Happy coding,
shuLhan