FWNX- build a full binary tree via an arrary- java version-优快云博客

本文介绍了一个使用给定数组构建二叉树的方法，并通过广度优先搜索(BFS)实现节点遍历的过程。文章详细展示了如何计算树的层数、节点的父节点以及构建过程中的关键步骤。

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package test_kepler;

public class TreefromA {
	
	int totalNum ;
	Node[] treeA;
	int[] array1 = new int[]{1,2,34,45,90};
	class Node
	{
		int value;
		Node leftChild ;
		Node rightChild;
		public boolean isVisted;
		public Node()
		{
			value =0;
			leftChild=null;
			rightChild=null;
			isVisted = false;
		}
		public void setLeftChild(Node lc)
		{
			this.leftChild = lc;
		}
		public void setrightChild(Node rc)
		{
			this.rightChild = rc;
		}
		public void print()
		{
			System.out.print(this.value +" ~~ ");
		}
	};
	public TreefromA(int [] a)
	{
		totalNum = a.length;
		treeA =  new Node[totalNum];
		for(int i = 0;i<totalNum;++i)
		{
			treeA[i] = new Node();
			System.out.println(a[i]+"==>");
			treeA[i].value = a[i];
		}
		buildTree();
		BFS();
	}
	//return is the whole layer - 1;
	public int getLayer(int array_length)
	{
		
		int i = 1;
		while(array_length>0)
		{
			array_length = array_length<<i;
			++i;
		}
		return i;		
	}
	//get ceiling 
	int getCeiling(double i)
	{
	    int result = (int)i;
	    int k = result +1;
	    System.out.println(i+"-->"+k);
	    return (result +1);
	}
	//get parentnum
	int getParent(int i)
	{
		if(i <=0) {System.out.println("this is wrong for finding father");return -1;}
		double ik = i/2;
		return getCeiling(ik)-1;
	}
	int power(int base,int i)
	{
		if(i == 0) return 1;
		else
		{
			return base * power(base,i-1);
		}
	}
	//!!!!
	int power2(int base,int i)
	{
		int result = 1;
		if(i ==0) return 1;
		else
		{
			for(int j = 0;j<i;++j)
			{
				result = result*base;
			}
			System.out.println("power2 = "+result);
			return result;
		}
	}
	public void buildTree()
	{
		int runtime = 0;
		for(int i = 1;i <= getLayer(totalNum)-1;++i)
		{
			for(int j = power(2,i)-1;j<=(power(2,i)-1)*2;++j)
			{
				System.out.println("j = "+j);
				if(j%2 ==1)
				{
					int f =getParent(j);
					System.out.println("f = "+f);
					treeA[f].print();
					treeA[getParent(j)].setLeftChild(treeA[j]);
				}
				else
				{
					treeA[getParent(j)].setrightChild(treeA[j]);
				}
				
				runtime++;
				
				if(runtime == totalNum-1)
				{
					return ;
				}
			}
		}
	}
	//TEST FUNCTION
	void BFS()
	{
		 qbytwostc testq = new qbytwostc<Node>();
		 testq.enQueue(treeA[0]);
		 while(!testq.isEmpty())
		 {
			 Node head = (Node) testq.dequeue();
			 if(head.isVisted == false)
			 head.print();
			 
			 
			 
			 if(head.rightChild!=null && head.rightChild.isVisted == false)
			 {
				 testq.enQueue(head.rightChild);
			 }
			 
			 if(head.leftChild!=null && head.leftChild.isVisted == false)
			 {
				 testq.enQueue(head.leftChild);
			 }
		 }
	}
	public static void main(String[] args) {
		
		//int a[] = {1,2,34,45,90};
		int[] array = new int[]{1,2,34,45,90,9090};
		TreefromA ta = new TreefromA(array);
		/*
		for(int i = 0;i<10;++i)
		{
			System.out.println(i+"->"+ta.power(2, i)+" --->>> "+ ta.power2(2, i));
		}*/
	}

}