java语言实现的二叉树的各种操作(包括递归与非递归遍历二叉树,求二叉树的高度，节点总数，叶子节点等)

 

/*java语言实现的二叉树的各种操作(包括递归与非递归遍历二叉树,求二叉树的高度，节点总数，叶子节点等)*/

BiTree.java

import java.util.Stack;public class BiTree {private BiTree leftTree;// 左子树private BiTree rightTree;// 右子树private Object data;// 节点数据public final static int MAX = 40;BiTree[] elements = new BiTree[MAX];// 层次遍历时保存各个节点int front;// 层次遍历时队首int rear;// 层次遍历时队尾// 构造函数public BiTree() {// TODO Auto-generated constructor stub}public BiTree(Object data) {this.data = data;}public BiTree(Object data, BiTree lefTree, BiTree righTree) {this.data = data;this.leftTree = lefTree;this.rightTree = righTree;}// 递归前序遍历二叉树public void preOrder(BiTree tree) {if (tree != null) {visit(tree.data);preOrder(tree.leftTree);preOrder(tree.rightTree);}}// 非递归实现前序遍历public void iterativePreOrder(BiTree tree) {Stack<BiTree> stack = new Stack<BiTree>();if (tree != null) {stack.push(tree);while (!stack.isEmpty()) {tree = stack.pop();visit(tree.data);if (tree.rightTree != null)stack.push(tree.rightTree);if (tree.leftTree != null)stack.push(tree.leftTree);}}}// 递归中序遍历二叉树public void inOrder(BiTree tree) {if (tree != null) {inOrder(tree.leftTree);visit(tree.data);inOrder(tree.rightTree);}}// 非递归实现中序遍历public void iterativeInOrder(BiTree tree) {Stack<BiTree> stack = new Stack<BiTree>();while (tree != null) {while (tree != null) {if (tree.rightTree != null)stack.push(tree.rightTree);// 当前节点右子入栈stack.push(tree);// 当前节点入栈tree = tree.leftTree;}tree = stack.pop();while (!stack.empty() && tree.rightTree == null) {visit(tree.data);tree = stack.pop();}visit(tree.data);if (!stack.empty())tree = stack.pop();elsetree = null;}}// 递归后序遍历二叉树public void postOrder(BiTree tree) {if (tree != null) {postOrder(tree.leftTree);postOrder(tree.rightTree);visit(tree.data);}}// 非递归实现后序遍历public void iterativePostOrder(BiTree tree) {BiTree tempTree = tree;Stack<BiTree> stack = new Stack<BiTree>();while (tree != null) {// 左子树入栈for (; tree.leftTree != null; tree = tree.leftTree)stack.push(tree);// 当前节点无右子或右子已经输出while (tree != null&& (tree.rightTree == null || tree.rightTree == tempTree)) {visit(tree.data);tempTree = tree;// 记录上一个已输出节点if (stack.empty())return;tree = stack.pop();}// 处理右子stack.push(tree);tree = tree.rightTree;}}// 层次遍历二叉树public void LayerOrder(BiTree tree) {elements[0] = tree;front = 0;rear = 1;while (front < rear) {try {if (elements[front].data != null) {System.out.print(elements[front].data + " ");if (elements[front].leftTree != null)elements[rear++] = elements[front].leftTree;if (elements[front].rightTree != null)elements[rear++] = elements[front].rightTree;front++;}} catch (Exception e) {break;}}}// 求二叉树的高度public static int height(BiTree tree) {if (tree == null)return 0;else {int leftTreeHeight = height(tree.leftTree);int rightTreeHeight = height(tree.rightTree);return leftTreeHeight > rightTreeHeight ? leftTreeHeight + 1: rightTreeHeight + 1;}}// 求data所对应结点的层数，如果对象不在树中,结果返回-1;否则结果返回该对象在树中所处的层次,规定根节点为第一层public int level(Object data) {int leftLevel, rightLevel;if (this == null)return -1;if (data == this.data)return 1;leftLevel = leftTree == null ? -1 : leftTree.level(data);rightLevel = rightTree == null ? -1 : rightTree.level(data);if (leftLevel < 0 && rightLevel < 0)return -1;return leftLevel > rightLevel ? leftLevel + 1 : rightLevel + 1;}// 求二叉树的结点总数public static int nodes(BiTree tree) {if (tree == null)return 0;else {int left = nodes(tree.leftTree);int right = nodes(tree.rightTree);return left + right + 1;}}// 求二叉树叶子节点的总数public static int leaf(BiTree tree) {if (tree == null)return 0;else {int left = leaf(tree.leftTree);int right = leaf(tree.rightTree);if (tree.leftTree == null && tree.rightTree == null)return left + right + 1;elsereturn left + right;}}// 求二叉树父节点个数public static int fatherNodes(BiTree tree) {if (tree == null || (tree.leftTree == null && tree.rightTree == null))return 0;else {int left = fatherNodes(tree.leftTree);int right = fatherNodes(tree.rightTree);return left + right + 1;}}// 求只有一个孩子结点的父节点个数public static int oneChildFather(BiTree tree) {int left, right;if (tree == null || (tree.rightTree == null && tree.leftTree == null))return 0;else {left = oneChildFather(tree.leftTree);right = oneChildFather(tree.rightTree);if ((tree.leftTree != null && tree.rightTree == null)|| (tree.leftTree == null && tree.rightTree != null))return left + right + 1;elsereturn left + right;/* 加1是因为要算上根节点 */}}// 求二叉树只拥有左孩子的父节点总数public static int leftChildFather(BiTree tree) {if (tree == null)return 0;else {int left = leftChildFather(tree.leftTree);int right = leftChildFather(tree.rightTree);if ((tree.leftTree != null && tree.rightTree == null))return left + right + 1;elsereturn left + right;}}// 求二叉树只拥有右孩子的结点总数public static int rightChildFather(BiTree tree) {if (tree == null || tree.rightTree == null)return 0;else {int left = rightChildFather(tree.leftTree);int right = rightChildFather(tree.rightTree);if (tree.leftTree == null && tree.rightTree != null)return left + right + 1;elsereturn left + right;}}// 计算有两个节点的父节点的个数public static int doubleChildFather(BiTree tree) {int left, right;if (tree == null)return 0;else {left = doubleChildFather(tree.leftTree);right = doubleChildFather(tree.rightTree);if (tree.leftTree != null && tree.rightTree != null)return (left + right + 1);/* 加1是因为要算上根节点 */elsereturn (left + right);}}// 访问根节点public void visit(Object data) {System.out.print(data + " ");}// 将树中的每个节点的孩子对换位置public void exChange() {if (this == null)return;if (leftTree != null)leftTree.exChange();if (rightTree != null)rightTree.exChange();BiTree temp = leftTree;leftTree = rightTree;rightTree = temp;}//递归求所有结点的和public static int getSumByRecursion(BiTree tree){if(tree==null){return 0;}else{int left=getSumByRecursion(tree.leftTree);int right=getSumByRecursion(tree.rightTree);return Integer.parseInt(tree.data.toString())+left+right;}}//非递归求二叉树中所有结点的和public static int getSumByNoRecursion(BiTree tree){Stack<BiTree> stack = new Stack<BiTree>();int sum=0;if(tree!=null){stack.push(tree);while(!stack.isEmpty()){tree=stack.pop();sum+=Integer.parseInt(tree.data.toString());if(tree.leftTree!=null) stack.push(tree.leftTree);if(tree.rightTree!=null)stack.push(tree.rightTree);}}return sum;}/** * @param args */public static void main(String[] args) {// TODO Auto-generated method stubBiTree e = new BiTree(5);BiTree g = new BiTree(7);BiTree h = new BiTree(8);BiTree l = new BiTree(12);BiTree m = new BiTree(13);BiTree n = new BiTree(14);BiTree k = new BiTree(11, n, null);BiTree j = new BiTree(10, l, m);BiTree i = new BiTree(9, j, k);BiTree d = new BiTree(4, null, g);BiTree f = new BiTree(6, h, i);BiTree b = new BiTree(2, d, e);BiTree c = new BiTree(3, f, null);BiTree tree = new BiTree(1, b, c);System.out.println("递归前序遍历二叉树结果: ");tree.preOrder(tree);System.out.println();System.out.println("非递归前序遍历二叉树结果: ");tree.iterativePreOrder(tree);System.out.println();System.out.println("递归中序遍历二叉树的结果为:");tree.inOrder(tree);System.out.println();System.out.println("非递归中序遍历二叉树的结果为:");tree.iterativeInOrder(tree);System.out.println();System.out.println("递归后序遍历二叉树的结果为:");tree.postOrder(tree);System.out.println();System.out.println("非递归后序遍历二叉树的结果为:");tree.iterativePostOrder(tree);System.out.println();System.out.println("层次遍历二叉树结果: ");tree.LayerOrder(tree);System.out.println();System.out.println("递归求二叉树中所有结点的和为:"+getSumByRecursion(tree));System.out.println("非递归求二叉树中所有结点的和为:"+getSumByNoRecursion(tree));System.out.println("二叉树中,每个节点所在的层数为:");for (int p = 1; p <= 14; p++)System.out.println(p + "所在的层为:" + tree.level(p));System.out.println("二叉树的高度为:" + height(tree));System.out.println("二叉树中节点总数为:" + nodes(tree));System.out.println("二叉树中叶子节点总数为:" + leaf(tree));System.out.println("二叉树中父节点总数为:" + fatherNodes(tree));System.out.println("二叉树中只拥有一个孩子的父节点数:" + oneChildFather(tree));System.out.println("二叉树中只拥有左孩子的父节点总数:" + leftChildFather(tree));System.out.println("二叉树中只拥有右孩子的父节点总数:" + rightChildFather(tree));System.out.println("二叉树中同时拥有两个孩子的父节点个数为:" + doubleChildFather(tree));System.out.println("--------------------------------------");tree.exChange();System.out.println("交换每个节点的左右孩子节点后......");System.out.println("递归前序遍历二叉树结果: ");tree.preOrder(tree);System.out.println();System.out.println("非递归前序遍历二叉树结果: ");tree.iterativePreOrder(tree);System.out.println();System.out.println("递归中序遍历二叉树的结果为:");tree.inOrder(tree);System.out.println();System.out.println("非递归中序遍历二叉树的结果为:");tree.iterativeInOrder(tree);System.out.println();System.out.println("递归后序遍历二叉树的结果为:");tree.postOrder(tree);System.out.println();System.out.println("非递归后序遍历二叉树的结果为:");tree.iterativePostOrder(tree);System.out.println();System.out.println("层次遍历二叉树结果: ");tree.LayerOrder(tree);System.out.println(); System.out.println("递归求二叉树中所有结点的和为:"+getSumByRecursion(tree));System.out.println("非递归求二叉树中所有结点的和为:"+getSumByNoRecursion(tree));System.out.println("二叉树中,每个节点所在的层数为:");for (int p = 1; p <= 14; p++)System.out.println(p + "所在的层为:" + tree.level(p));System.out.println("二叉树的高度为:" + height(tree));System.out.println("二叉树中节点总数为:" + nodes(tree));System.out.println("二叉树中叶子节点总数为:" + leaf(tree));System.out.println("二叉树中父节点总数为:" + fatherNodes(tree));System.out.println("二叉树中只拥有一个孩子的父节点数:" + oneChildFather(tree));System.out.println("二叉树中只拥有左孩子的父节点总数:" + leftChildFather(tree));System.out.println("二叉树中只拥有右孩子的父节点总数:" + rightChildFather(tree));System.out.println("二叉树中同时拥有两个孩子的父节点个数为:" + doubleChildFather(tree));}}