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); } }
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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