Determine whether a binary tree is a sub-structure of another binary tree

Determine whether a binary tree is a sub-structure of another binary tree

1. Question description:

How can we determine whether a binary tree is another sub-structure?
For example:

2

/\

9 8

/\/

2 3 5

/

6

The sub-structure is
9
/\
2 3

2,Analyze the problem:

Algorithms related to binary trees can be solved through recursion. To write a correct recursive program, you must first analyze the conditions for correct recursive termination.

Let's talk about the time when recursion ends.

<1> If root2 of the second binary tree is null, it indicates that root2 is the sub-structure of root1 of the first Binary Tree, and true is returned.

<2> when root1 is empty, root2 is not empty yet. This indicates that root2 is not a sub-structure of root1 and false is returned.

<3> there are two ideas for recursion:

Method 1: Find the node with the same node value as the value of root2 in root1. If the node value is found, determine whether root2 is the sub-structure starting with this node. Therefore, the issubtree () is used to determine.

Method 2: directly determine whether or not the sub-tree of root2 is also a corresponding sub-structure. If the values are different, the left and right subtree of root1. Pay special attention to the recursive logic judgment of the last two sentences.

3. Code:

Method 1:

// Determine whether root2 is a sub-structure starting with root1.

Boolissubtree (bitreenode * root1, bitreenode * root2)

{

// Determine root2 first

If (root2 = NULL)

Return true;

If (root1 = NULL)

Return false;

If (root1-> data! = Root2-> data)

Return false;

Returnissubtree (root1-> LC, root2-> Lc) & issubtree (root1-> RC, root2-> RC );

}

// Recursively search for the same value as root2 in the tree with root1 as the node. If so, call issubtree (root1, root2 );

Boolcheckifsubtree (bitreenode * root1, bitreenode * root2)

{

If (root1 = NULL)

Return false;

Bool result = false;

If (root1-> DATA = root2-> data)

Result = issubtree (root1, root2 );

If (result = false)

Result = checkifsubtree (root1-> LC, root2 );

If (result = false)

Result = checkifsubtree (root1-> RC, root2 );

Return result;

}

Method 2:

// Recursion

Boolcheckifsubtree2 (bitreenode * root1, bitreenode * root2)

{

If (root2 = NULL)

Return true;

If (root1 = NULL)

Return false;

If (root1-> DATA = root2-> data)

Returncheckifsubtree2 (root1-> LC, root2-> Lc) & checkifsubtree2 (root1-> RC, root2-> RC );

Else

Returncheckifsubtree2 (root1-> LC, root2) | checkifsubtree2 (root1-> RC, root2 );

}

4. Code for generating Binary Trees in FIFO order

Struct bitreenode
{
Struct bitreenode * LC;
Struct bitreenode * RC;
Int data;
};

Void createbitree (bitreenode * & root)
{
Int ch;
Cin> CH;
If (CH = 0)
{
Root = NULL;
}
Else
{
Root = (bitreenode *) malloc (sizeof (bitreenode ));
If (root = NULL)
Exit (1 );
Root-> DATA = CH;
// Cout <"--" <root-> data <Endl;
Createbitree (root-> Lc );
Createbitree (root-> RC );
}
}

5. Summary

Be sure to understand recursion well, especially when recursion ends. This is the key to solving most recursive problems.

This code has been debugged in VC ++ 6.0. You are welcome to raise your question.