How to build a binary tree using java

Directory:

1. Put a Assign the value of array to a binary tree

2. Specific code

Note:

1. The subscript of the parent node array ranges from 0 to n/2 -1, but it should be less than n/2-1 when traversing, because the last The parent node may not have a right child. When n/2-1 is an odd number, it has a right child, and when it is an even number, it only has a left child.

##2. Node. The subscript of the left child is 2n+1, and the subscript of the right child is 2n+2

##1. Tree construction method

2. Specific code

package tree;  
  
import java.util.LinkedList;  
import java.util.List;  
  
/** 
 * 功能：把一个数组的值存入二叉树中，然后进行3种方式的遍历 
 *  
 * 参考资料0:数据结构(C语言版)严蔚敏 
 *  
 * 参考资料1：http://zhidao.baidu.com/question/81938912.html 
 *  
 * 参考资料2：http://cslibrary.stanford.edu/110/BinaryTrees.html#java 
 *  
 * @author ocaicai@yeah.net @date: 2011-5-17 
 *  
 */  
public class BinTreeTraverse2 {  
  
    private int[] array = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };  
    private static List<Node> nodeList = null;  
  
    /** 
     * 内部类：节点 
     *  
     * @author ocaicai@yeah.net @date: 2011-5-17 
     *  
     */  
    private static class Node {  
        Node leftChild;  
        Node rightChild;  
        int data;  
  
        Node(int newData) {  
            leftChild = null;  
            rightChild = null;  
            data = newData;  
        }  
    }  
  
    public void createBinTree() {  
        nodeList = new LinkedList<Node>();  
        // 将一个数组的值依次转换为Node节点  
        for (int nodeIndex = 0; nodeIndex < array.length; nodeIndex++) {  
            nodeList.add(new Node(array[nodeIndex]));  
        }  
        // 对前lastParentIndex-1个父节点按照父节点与孩子节点的数字关系建立二叉树  
        for (int parentIndex = 0; parentIndex < array.length / 2 - 1; parentIndex++) {  
            // 左孩子  
            nodeList.get(parentIndex).leftChild = nodeList  
                    .get(parentIndex * 2 + 1);  
            // 右孩子  
            nodeList.get(parentIndex).rightChild = nodeList  
                    .get(parentIndex * 2 + 2);  
        }  
        // 最后一个父节点:因为最后一个父节点可能没有右孩子，所以单独拿出来处理  
        int lastParentIndex = array.length / 2 - 1;  
        // 左孩子  
        nodeList.get(lastParentIndex).leftChild = nodeList  
                .get(lastParentIndex * 2 + 1);  
        // 右孩子,如果数组的长度为奇数才建立右孩子  
        if (array.length % 2 == 1) {  
            nodeList.get(lastParentIndex).rightChild = nodeList  
                    .get(lastParentIndex * 2 + 2);  
        }  
    }  
  
    /** 
     * 先序遍历 
     *  
     * 这三种不同的遍历结构都是一样的，只是先后顺序不一样而已 
     *  
     * @param node 
     *            遍历的节点 
     */  
    public static void preOrderTraverse(Node node) {  
        if (node == null)  
            return;  
        System.out.print(node.data + " ");  
        preOrderTraverse(node.leftChild);  
        preOrderTraverse(node.rightChild);  
    }  
  
    /** 
     * 中序遍历 
     *  
     * 这三种不同的遍历结构都是一样的，只是先后顺序不一样而已 
     *  
     * @param node 
     *            遍历的节点 
     */  
    public static void inOrderTraverse(Node node) {  
        if (node == null)  
            return;  
        inOrderTraverse(node.leftChild);  
        System.out.print(node.data + " ");  
        inOrderTraverse(node.rightChild);  
    }  
  
    /** 
     * 后序遍历 
     *  
     * 这三种不同的遍历结构都是一样的，只是先后顺序不一样而已 
     *  
     * @param node 
     *            遍历的节点 
     */  
    public static void postOrderTraverse(Node node) {  
        if (node == null)  
            return;  
        postOrderTraverse(node.leftChild);  
        postOrderTraverse(node.rightChild);  
        System.out.print(node.data + " ");  
    }  
  
    public static void main(String[] args) {  
        BinTreeTraverse2 binTree = new BinTreeTraverse2();  
        binTree.createBinTree();  
        // nodeList中第0个索引处的值即为根节点  
        Node root = nodeList.get(0);  
  
        System.out.println("先序遍历：");  
        preOrderTraverse(root);  
        System.out.println();  
  
        System.out.println("中序遍历：");  
        inOrderTraverse(root);  
        System.out.println();  
  
        System.out.println("后序遍历：");  
        postOrderTraverse(root);  
    }  
  
}

Output result:

先序遍历：  
1 2 4 8 9 5 3 6 7   
中序遍历：  
8 4 9 2 5 1 6 3 7   
后序遍历：  
8 9 4 5 2 6 7 3 1

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