遍历递归树求递推数列通项

考虑K阶变系数线性递推方程：

现给定初值a1,a2,—,ak和n>k,要求编程打印an,an-1,—,ak+1的值

该问题用常规的迭代法非常容易解决，现在要考虑的是用遍历递归调用树的方法求解。n=7,k=3时，递归调用树为

package programm;           //遍历递归树中所有节点，相同值会重复计算
import java.util.*;public class RecursionTree
{public static void main(String[] args){final int K=3;final int N=7;int[] initial=new int[K];for (int i=0; i<K; i++)initial[i]=i+1;Stack<Node> stack=new Stack<Node>();stack.push(new Node(0, 0, N));boolean TF=true;int sign=0, n=stack.getTop().getIndex();while(!stack.isEmpty()){if (1<=n&&n<=K){stack.getTop().setSum(stack.getTop().getSum()+initial[n-1]*(stack.getTop().getIndex()+stack.getTop().getFlag()));TF=false;n=stack.getTop().getIndex();}else{ if (TF==false){stack.getTop().setFlag(stack.getTop().getFlag()+1);if (stack.getTop().getFlag()>K){Node temp=stack.pop();temp.setSum(temp.getSum()+temp.getIndex());if (stack.isEmpty()){System.out.println("The "+temp.getIndex()+"nd item is "+temp.getSum());}else{                if (stack.getTop().getFlag()==1 && temp.getIndex()>sign){sign=temp.getIndex();System.out.println("The "+temp.getIndex()+"nd item is "+temp.getSum());}stack.getTop().setSum(stack.getTop().getSum()+temp.getSum()*(stack.getTop().getIndex()+stack.getTop().getFlag()));n=stack.getTop().getIndex();TF=false;}}else{if ((n=stack.getTop().getIndex()-stack.getTop().getFlag())>K){stack.push(new Node(0, 0, n));TF=true;}else{continue;}}}else{stack.getTop().setFlag(stack.getTop().getFlag()+1);if ((n=stack.getTop().getIndex()-1)>K){stack.push(new Node(0, 0, n));TF=true;}else{continue;}}}}}}class Node
{private int sum;private int flag;private int index;    public Node(int sum, int flag, int index){this.sum=sum;this.flag=flag;this.index=index;}public int getSum(){return sum;}public void setSum(int sum){this.sum=sum;}public int getFlag(){return flag;}public void setFlag(int flag){this.flag=flag;}public int getIndex(){return index;}
}class Stack<E>
{ArrayList<E> list=new ArrayList<E>();public boolean isEmpty(){return list.isEmpty();    }public int getSize(){return list.size();}public E getTop(){return list.get(getSize()-1);}public E pop(){E interval=list.get(getSize()-1);list.remove(getSize()-1);return interval;}public void push(E Ob){list.add(Ob);}
}

运行结果：

以上方法的问题是，相同的项会被重复计算多次，如a4被重复计算了三次，所以以上方法需要改进。用数组保存已求出项的值，遍历
到相同项时直接从数组中找到并使用其值即可。要实现这一点，需要修改源程序41行，48行，60-61行。修改后的代码如下：
程序二：

package programm;   //不遍历递归树中所有节点,相同值不会重复计算
import java.util.*;public class RecursionTree
{public static void main(String[] args){final int K=3;   //递推方程阶数final int N=7;   //要求的部分数列中下标最大的项的下标int[] initial=new int[K];   //初值a1----akfor (int i=0; i<K; i++)    //令初值a1----ak分别为1,2---,kinitial[i]=i+1;Stack<Node> stack=new Stack<Node>();   //建立Node类型栈对象stack.push(new Node(0, 0, N));     //递归树根节点压入栈，该节点flag=sum=0，index=Nboolean TF=true;    //true向前进至下一层,false回溯至上一层int sign=0, n=stack.getTop().getIndex();   //sign记录当前已遍历节点中下标最大的节点,n初始化为根节点下标int[] result=new int[N-K];   //记录ak+1---aN项值的数组while(!stack.isEmpty())   //栈不空{if (1<=n&&n<=K)  //到达递归树中可直接写出值的叶子结点{stack.getTop().setSum(stack.getTop().getSum()+initial[n-1]*(stack.getTop().getIndex()+stack.getTop().getFlag()));  //将叶子结点值乘以系数累加至父节点TF=false;n=stack.getTop().getIndex();    //回溯}else{ if (TF==false){stack.getTop().setFlag(stack.getTop().getFlag()+1);   //当前节点flag增一，试探下一子节点if (stack.getTop().getFlag()>K)    //当前节点所有子节点均试探完毕{Node temp=stack.pop();   //从栈中弹出当前节点temp.setSum(temp.getSum()+temp.getIndex());   //加上尾数f(n),得到当前节点的值if (stack.isEmpty())  //栈空,temp引用根节点，根节点值已求出，存放在temp的sum中{result[N-K-1]=temp.getSum();   //在数组result中存放根节点即aN的值System.out.println("The "+temp.getIndex()+"nd item is "+temp.getSum());  //打印aN的值}else{                if (stack.getTop().getFlag()==1 && temp.getIndex()>sign)  //找到当前已遍历节点中下标最大节点{sign=temp.getIndex();   //设置新的最大下标result[temp.getIndex()-K-1]=temp.getSum();  //当前节点值存放至result数组中System.out.println("The "+temp.getIndex()+"nd item is "+temp.getSum());  //打印当前节点值}stack.getTop().setSum(stack.getTop().getSum()+temp.getSum()*(stack.getTop().getIndex()+stack.getTop().getFlag()));  //当前节点值乘以系数累加至父节点n=stack.getTop().getIndex();TF=false;  //回溯}}else{if ((n=stack.getTop().getIndex()-stack.getTop().getFlag())>K)  //前进至非叶子结点{stack.getTop().setSum(stack.getTop().getSum()+result[n-K-1]*(stack.getTop().getIndex()+stack.getTop().getFlag()));  //当前节点值之前已算出，从数组result中找出该值乘以系数累加至父节点n=stack.getTop().getIndex();TF=false;      //回溯}else{continue;  //直接进至下一轮循环，累加叶子节点值和系数乘积至父节点}}}else{stack.getTop().setFlag(stack.getTop().getFlag()+1); //进至新的一层,当前节点flag增一试探下一层节点if ((n=stack.getTop().getIndex()-1)>K) //下一层第一个节点非叶子{stack.push(new Node(0, 0, n));    //该节点压入栈TF=true;  }                                      else{continue;    //同上}}}}}}class Node    //递归树节点类
{private int sum;     //当前节点值private int flag;    //当前节点的父节点下标和当前节点下标之差private int index;    //当前节点下标public Node(int sum, int flag, int index)   //构造方法{this.sum=sum;this.flag=flag;   this.index=index;}public int getSum(){return sum;}public void setSum(int sum){this.sum=sum;}//访问器和修改器public int getFlag(){return flag;}public void setFlag(int flag){this.flag=flag;}public int getIndex(){return index;}
}class Stack<E>            //泛型栈类
{ArrayList<E> list=new ArrayList<E>();public boolean isEmpty(){return list.isEmpty();    }public int getSize(){return list.size();}public E getTop(){return list.get(getSize()-1);}public E pop(){E interval=list.get(getSize()-1);list.remove(getSize()-1);return interval;}public void push(E Ob){list.add(Ob);}
}

可以验证运行结果和程序一相同，但是节省了运行时间
删掉程序一37行，修改24行和51行可以得到程序三，用于计算递推数列an=an-1+—an-kn>k a1=1—ak=k当给定n>k时an,—,ak+1的值。
程序三(n=7, k=2, a1=1, a2=2)：

package programm;          //an=an-1+---an-k型递归式求解，遍历所有节点
import java.util.*;public class RecursionTree
{public static void main(String[] args){final int K=2;final int N=7;int[] initial=new int[K];for (int i=0; i<K; i++)initial[i]=i+1;Stack<Node> stack=new Stack<Node>();stack.push(new Node(0, 0, N));boolean TF=true;int sign=0, n=stack.getTop().getIndex();while(!stack.isEmpty()){if (1<=n&&n<=K){stack.getTop().setSum(stack.getTop().getSum()+initial[n-1]);TF=false;n=stack.getTop().getIndex();}else{ if (TF==false){stack.getTop().setFlag(stack.getTop().getFlag()+1);if (stack.getTop().getFlag()>K){Node temp=stack.pop();if (stack.isEmpty()){System.out.println("The "+temp.getIndex()+"nd item is "+temp.getSum());}else{                if (stack.getTop().getFlag()==1 && temp.getIndex()>sign){sign=temp.getIndex();System.out.println("The "+temp.getIndex()+"nd item is "+temp.getSum());}stack.getTop().setSum(stack.getTop().getSum()+temp.getSum());n=stack.getTop().getIndex();TF=false;}}else{if ((n=stack.getTop().getIndex()-stack.getTop().getFlag())>K){stack.push(new Node(0, 0, n));TF=true;}else{continue;}}}else{stack.getTop().setFlag(stack.getTop().getFlag()+1);if ((n=stack.getTop().getIndex()-1)>K){stack.push(new Node(0, 0, n));TF=true;}else{continue;}}}}}}class Node
{private int sum;private int flag;private int index;    public Node(int sum, int flag, int index){this.sum=sum;this.flag=flag;this.index=index;}public int getSum(){return sum;}public void setSum(int sum){this.sum=sum;}public int getFlag(){return flag;}public void setFlag(int flag){this.flag=flag;}public int getIndex(){return index;}
}class Stack<E>
{ArrayList<E> list=new ArrayList<E>();public boolean isEmpty(){return list.isEmpty();    }public int getSize(){return list.size();}public E getTop(){return list.get(getSize()-1);}public E pop(){E interval=list.get(getSize()-1);list.remove(getSize()-1);return interval;}public void push(E Ob){list.add(Ob);}
}

运行结果：

显然，所求得的即为斐波那契数列的第4-8项
为了便于读者理解程序一二三，下面给出一个经过简化的程序四，实现思想和程序一二三基本相同且功能和程序三相同
程序四（n=7, k=2, a1=1, a2=2）:不难验证运行结果和程序三是一致的

package RecursionTree;    //简化版的an=an-1+---an-k型递归式求解，遍历所有节点
import java.util.*;public class recursiontree
{public static void main(String[] args) {final int K=2;final int N=7;int[] initial=new int[K];for (int i=0; i<K; i++)initial[i]=i+1;Stack<Node> stack=new Stack<Node>();stack.push(new Node(0, N));boolean TF=true;int sum=0;while (!stack.isEmpty()){if (1<=stack.getTop().getIndex() && stack.getTop().getIndex()<=K){sum+=initial[stack.getTop().getIndex()-1];stack.pop();TF=false;}else{if (TF==false){stack.getTop().setFlag(stack.getTop().getFlag()+1);if (stack.getTop().getFlag()>K){stack.pop();TF=false;}else{stack.push(new Node(0, stack.getTop().getIndex()-stack.getTop().getFlag()));TF=true;}}else{stack.getTop().setFlag(stack.getTop().getFlag()+1);stack.push(new Node(0, stack.getTop().getIndex()-1));TF=true;}}}System.out.println("The "+N+"nd item is "+sum);}
}class Node
{private int flag;private int index;    public Node(int flag, int index){this.flag=flag;this.index=index;}    public int getFlag(){return flag;}public void setFlag(int flag){this.flag=flag;}public int getIndex(){return index;}
}class Stack<E>
{ArrayList<E> list=new ArrayList<E>();public boolean isEmpty(){return list.isEmpty();    }public int getSize(){return list.size();}public E getTop(){return list.get(getSize()-1);}public E pop(){E interval=list.get(getSize()-1);list.remove(getSize()-1);return interval;}public void push(E Ob){list.add(Ob);}
}

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