众所周知，任何有序的林都可以用唯一的二叉树来表示。 任何人都可以帮我找到一个确定二叉树修剪数的算法吗？
答案 0 :(得分：2)
If a binary tree is represented in L and R array, e.g.
The tree would be represented then
1 L[1]=2, R[1]=4
/ \ L[2]=0, R[2]=3
2 4 L[3]=0, R[3]=0
\ L[4]=0, R[4]=0
3
Then this algorithm will help you in getting the pruning number:
public int getPruningNumber()
{
int index=1;
int l=0;
if(L[index-1]!=0)
{
l=getPruningNumberRecursilvely(index);
//System.out.println("l="+l);
}
int r=0;
boolean allRightSubTreeHaveSamePruningNo=true;
while(R[index-1]!=0)
{
index=R[index-1];
int k=getPruningNumberRecursilvely(index);
if(r==0)
r=k;
else if(k>r)
{
r=k;
allRightSubTreeHaveSamePruningNo=false;
}
else if(k<r)
allRightSubTreeHaveSamePruningNo=false;
// System.out.println("k="+k);
}
if(allRightSubTreeHaveSamePruningNo&&r==l)
return l+1;
return r>l?r:l;
}
private int getPruningNumberRecursilvely(int index)
{
if(L[index-1]==0&&R[index-1]==0)
return 1;
int l=0,r=0;
if(L[index-1]!=0)
{
l=getPruningNumberRecursilvely(L[index-1]);
// System.out.println("in rec l::"+l+" for index:"+index);
boolean allRightSubTreeHaveSamePruningNo=true;
index=L[index-1];
while(index!=0&&R[index-1]!=0)
{
int k=getPruningNumberRecursilvely(R[index-1]);
index=R[R[index-1]-1];
if(r==0)
r=k;
else if(k>r)
{
r=k;
allRightSubTreeHaveSamePruningNo=false;
}
// System.out.println("in rec k::"+" for index:"+index);
}
if(allRightSubTreeHaveSamePruningNo&&r==l)
return l+1;
return r>l?r:l;
}
return 1;
}
答案 1 :(得分：2)
1. Yes for the given case the pruning number will be 2 and i guess my program is also giving the same answer.
Given Binary tree:
1 1
/ / \
2 equivalent forest 2 3
\
3
2. All the right children of the root of binary tree is representing a tree in forest and so does the right of this right child of root. i.e.
1
/ \
2 5
/ \ / \
3 4 6 10
\
7
/ \
8 9
this binary tree is representing this forest.
1 5 10
/ \ / \ \
2 4 6 7 9
/ /
3 8
3. i m pasting the full code to compute the pruning number:-
public class PruningNumber {
private int L[]=null;
private int R[]=null;
public PruningNumber(int L[],int R[])
{
this.L=L;
this.R=R;
}
public int getPruningNumber()
{
int index=1;
int l=0;
if(L[index-1]!=0)
{
l=getPruningNumberRecursilvely(index);
}
int r=0;
boolean allRightSubTreeHaveSamePruningNo=true;
while(R[index-1]!=0)
{
index=R[index-1];
int k=getPruningNumberRecursilvely(index);
if(r==0)
r=k;
else if(k>r)
{
r=k;
allRightSubTreeHaveSamePruningNo=false;
}
else if(k<r)
allRightSubTreeHaveSamePruningNo=false;
}
if(allRightSubTreeHaveSamePruningNo&&r==l)
return l+1;
return r>l?r:l;
}
private int getPruningNumberRecursilvely(int index)
{
if(L[index-1]==0&&R[index-1]==0)
return 1;
int l=0,r=0;
if(L[index-1]!=0)
{
l=getPruningNumberRecursilvely(L[index-1]);
boolean allRightSubTreeHaveSamePruningNo=true;
index=L[index-1];
while(index!=0&&R[index-1]!=0)
{
int k=getPruningNumberRecursilvely(R[index-1]);
index=R[R[index-1]-1];
if(r==0)
r=k;
else if(k>r)
{
r=k;
allRightSubTreeHaveSamePruningNo=false;
}
}
if(allRightSubTreeHaveSamePruningNo&&r==l)
return l+1;
return r>l?r:l;
}
return 1;
}
public static void main(String args[])
{
int L[]={2,0,4,0};
int R[]={3,0,0,0};
System.out.println(new PruningNumber(L,R).getPruningNumber());
int L1[]={2,3,0,0,6,0,0};
int R1[]={5,4,0,0,7,0,0};
System.out.println(new PruningNumber(L1,R1).getPruningNumber());
//the case u r discussing
int L3[]={2,0,0};
int R3[]={0,3,0};
PruningNumber pruningNumber=new PruningNumber(L3,R3);
System.out.println(pruningNumber.getPruningNumber());
int L4[]={2 ,3,0,5,6,0,0,9,0,0, 12,13,0,15,0,17,18,0,0,21,0,0,0};
int R4[]={11,4,0,8,7,0,0,10,0,0,0,14,0,16,0,20,19,0,0,22,0,0,0};
pruningNumber=new PruningNumber(L4,R4);
System.out.println(pruningNumber.getPruningNumber());
int L5[]={2,3,0,0,6,0,0,0};
int R5[]={0,5,4,0,8,7,0,0};
pruningNumber=new PruningNumber(L5,R5);
System.out.println(pruningNumber.getPruningNumber());
int L6[]={2,3,4,0,0,0,0};
int R6[]={0,0,0,5,6,7,0};
pruningNumber=new PruningNumber(L6,R6);
System.out.println(pruningNumber.getPruningNumber());
}
}
答案 2 :(得分：1)
我对这个解决方案有点怀疑，特别是对于案例
int L[]={2,0,4,0};
int R[]={3,0,0,0};
为什么它的修剪数是2？
它的森林由
提供 1 && 3
/ /
2 4
所以它的修剪数应该是1.答案显示为2
答案 3 :(得分：0)
算法检查L [index-1] ... L [0]为0。 在上面的二叉树表示中，从L [1]开始（你的索引从1开始，而不是0）。
您的算法将始终返回1.
还要考虑这种情况：
L[1]=2, R[1]=0
L[2]=0, R[2]=3
L[3]=0, R[3]=0
这应该导致修剪数为2.原因是因为3在2的右边，这意味着在有序的森林2和3中是兄弟姐妹和1的孩子。细丝是2和如图3所示，将在第一次修剪中移除，然后在第二次修剪中移除1。