数组法构建二叉树
public class Main { static LinkedList<Tree> linkedList=new LinkedList<Tree>(); static int[] a= {1,2,3,4,5,6,7,8,9}; public static void main(String[] args) { createBinTree(); //获取链表的头结点 Tree root=linkedList.get(0); System.out.println("先序遍历："); preOrder(root); System.out.println(); System.out.println("中序遍历："); inOrder(root); System.out.println(); System.out.println("后序遍历："); postOrder(root); System.out.println(); } //用数组的方式构建二叉树 public static void createBinTree() { //把数组中的值转为二叉树的值并存入LinkedKist集合 for(int i=0;i<a.length;i++) { linkedList.add(new Tree(a[i])); } //把LinkedList集合转成二叉树的形式 for(int j=0;j<a.length/2-1;j++) { linkedList.get(j).leftTree=linkedList.get(j*2+1); linkedList.get(j).rightTree=linkedList.get(j*2+2); } //最后一个父节点单独计算 int lastTree=a.length/2-1; linkedList.get(lastTree).leftTree=linkedList.get(lastTree*2+1); //数组元素个数是奇数则最后一个父节点有右孩子 if(a.length%2==1) { linkedList.get(lastTree).rightTree=linkedList.get(lastTree*2+2); } } //先序遍历 public static void preOrder(Tree tree) { if(tree!=null) { System.out.print(tree.data); preOrder(tree.leftTree); preOrder(tree.rightTree); } } //中序遍历 public static void inOrder(Tree tree) { if(tree!=null) { inOrder(tree.leftTree); System.out.print(tree.data); inOrder(tree.rightTree); } } //后序遍历 public static void postOrder(Tree tree) { if(tree!=null) { postOrder(tree.leftTree); postOrder(tree.rightTree); System.out.print(tree.data); } } } class Tree{ int data; Tree leftTree; Tree rightTree; public Tree(int data) { this.data=data; }}
转载于:https://www.cnblogs.com/sgbe/p/11427856.html
相关资源：二叉树遍历的流程图(递归 非递归 前中后序)