creating red black tree
TRANSCRIPT
CREATING RED BLACK TREE
CREATE A RED BLACK TREE WITH THE FOLLOWING ELEMENTS:
10,5,60,25,1,23
Red black tree properties.
• Each node is marked either red or black.
• Root node is always black
• Leaf nodes (null pointers) are always black.
• There should not be two consecutive reds.
• Number of black nodes is same from root to any leaf node.
10,5,60,25,1,23
• INSERT 10
• INITIALLY TREE IS EMPTY. MAKE 10 AS ROOT AND MARK IT BLACK.
10
10,5,60,25,1,23
• INSERT 5
• 5<10 SO, GO TO LEFT SUBTREE OF 10, WHICH IS EMPTY SO INSERT 5 TO LEFT OF 10 AND MARK IT RED.
10
5
10,5,60,25,1,23
• INSERT 60
• 60>10 SO, GO TO RIGHT SUBTREE OF 10, WHICH IS EMPTY SO INSERT 60 TO RIGHT OF 10 AND MARK IT RED.
10
5 60
10,5,60,25,1,23
• INSERT 25
• 25>10 SO, GO TO RIGHT SUBTREE OF 10
• 25<60 , GO TO LEFT SUBTREE OF 60, WHICH IS EMPTY SO INSERT 25 TO LEFT OF 60 AND MARK IT RED.
It is of the form RL.
U = 25
P(u) = 60
G(u) = 10
S(p(u))=5 =>red
SINCE COLOR OF SIBLING OF P(U) is red. So recoloring will solve the problem.
5
10
60
25
RB TREE PROPERTY VIOLATED. NO CONSECUTIVE TWO REDS.
• Make p(u) and s(p(u)) black.
• Make g(u) red.
• But in this example g(u) is root. So we donot make it red.
5
10
60
25
10,5,60,25,1,23
• INSERT 1
• 1<10 SO, GO TO LEFT SUBTREE OF 10
• 1<5 , GO TO LEFT SUBTREE OF 5, WHICH IS EMPTY SO INSERT 1 TO LEFT OF 5 AND MARK IT RED.
5
10
60
251
10,5,60,25,1,23• INSERT 23
• 23>10 SO, GO TO RIGHT SUBTREE OF 10
• 23<60 , GO TO LEFT SUBTREE OF 60,
• 23<25, GO TO LEFT SUBTREE OF 25,
WHICH IS EMPTY SO INSERT 23 TO LEFT
OF 25 AND MARK IT RED.
RB TREE PROPERTY VIOLATED.(TWO CONSECUTIVE REDS (25 AND 23)
U=23
P(u)=25
G(u)=60
S(p(u))=nil (blank)
It is of the form LL.
U is left child of p(u) and p(u) is also left child of g(u).
• Since s(p(u)) is blank, Rotation is required.{make p(u) parent of u and g(u)
Make p(u) black and g(u) red.}
5
10
60
251
23
Final RB tree with elements: 10,5,60,25,1,23
5
10
1 60
25
23
