65 lines
2.1 KiB
Java
65 lines
2.1 KiB
Java
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/*This Java program is to Implement binary tree and check whether a tree is subtree of another. This can be done in two ways. A tree can be subtree of another if they have same structure (same object references but different value) and with same structure and values. This given class checks for both.*/
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//This is a java program to check whether a binary tree is subtree of another tree
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class Btrees
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{
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Object value;
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Btrees Left;
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Btrees Right;
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Btrees(int k)
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{
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value = k;
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}
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}
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public class SubBinaryTree
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{
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public static boolean ifsubtreeRef(Btrees t1, Btrees t2)
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{
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if (t2 == null)
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return true;
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if (t1 == null)
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return false;
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return (t1 == t2) || ifsubtreeRef(t1.Left, t2)
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|| ifsubtreeRef(t1.Right, t2);
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}
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public static boolean ifsubtreeValue(Btrees t1, Btrees t2)
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{
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if (t2 == null)
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return true;
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if (t1 == null)
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return false;
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if (t1.value == t2.value)
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if (ifsubtreeValue(t1.Left, t2.Left)
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&& ifsubtreeValue(t1.Right, t2.Right))
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return true;
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return ifsubtreeValue(t1.Left, t2) || ifsubtreeValue(t1.Right, t2);
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}
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public static void main(String[] args)
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{
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Btrees t1 = new Btrees(1);
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t1.Left = new Btrees(2);
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t1.Right = new Btrees(3);
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t1.Right.Left = new Btrees(4);
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t1.Right.Right = new Btrees(5);
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Btrees t2 = new Btrees(3);
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t2.Left = new Btrees(4);
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t2.Right = new Btrees(5);
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if (ifsubtreeRef(t1, t2))
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System.out.println("T2 is sub-tree of T1 (Reference wise)");
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else
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System.out.println("T2 is NOT sub-tree of T1 (Reference wise)");
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if (ifsubtreeValue(t1, t2))
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System.out.println("T2 is sub-tree of T1 (Value wise)");
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else
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System.out.println("T2 is NOT sub-tree of T1 (Value wise)");
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}
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}
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/*
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T2 is NOT sub-tree of T1 (Reference wise)
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T2 is sub-tree of T1 (Value wise)
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