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接着我们就要写一个比较复杂的数据结构的，但是这个数据结构是很重要的,假如你想深入的学习算法等等.我们来模拟一下二叉树。 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 private int front;//层次遍历时队首 private int rear;//层次遍历时队尾 //构造函数 public BiTree(){ } public BiTree(Object data) { this.data = data; } public BiTree(Object data, BiTree lefTree, BiTree righTree){ this.data = data; this.leftTree = lefTree; this.rightTree = righTree; } //递归遍历二叉树(前序) public void preOrder(BiTree tree) { if (tree != null) { visit(tree.data);//展示数据 preOrder(tree.leftTree); preOrder(tree.rightTree); } } //递归遍历二叉树(中序) public void inOrder(BiTree tree) { if (tree != null) { inOrder(tree.leftTree); //访问数据 visit(tree.data); inOrder(tree.rightTree); } } //递归遍历二叉树(后序) public void postOrder(BiTree tree) { if (tree != null) { postOrder(tree.leftTree); postOrder(tree.rightTree); visit(tree.data); } } //非递归实现遍历(前序) public void iterativePreOrder(BiTree tree) { //堆栈 Stack<BiTree> stack = new Stack<BiTree>(); if (tree != null) { stack.push(tree); while (!stack.isEmpty()) { //先访问根 tree = stack.pop(); //访问根节点 visit(tree.data); if (tree.rightTree != null) stack.push(tree.rightTree); if (tree.leftTree != null) stack.push(tree.leftTree); } } } //中序实现二叉树遍历(非递归) public void iterativeInOrder(BiTree tree){ //堆栈 Stack<BiTree> stack = new Stack<BiTree>(); while(tree!=null)//遍历所有的tree { while(tree!=null)//装入最左边的tree { if(tree.rightTree!=null) { stack.push(tree.rightTree); } stack.push(tree); //指向左tree tree=tree.leftTree; } //遍历树 tree=stack.pop(); //右树为空的时候，说明需要输出左边的树 while(!stack.empty() && tree.rightTree==null) { visit(tree.data); tree=stack.pop(); } //当左树遍历完成后要遍历根接点 visit(tree.data); //然后继续遍历 if(!stack.empty()) { tree=stack.pop(); } else { tree=null;//退出循环 } } } //非递归实现遍历(后续) public void iterativePostOrder(BiTree tree) { //临时变量 BiTree tempTree=tree; Stack<BiTree> stack = new Stack<BiTree>(); //遍历所有的tree while(tree!=null) { while(tree.leftTree!=null) { stack.push(tree);//装载左树 tree=tree.leftTree; } //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(); } //装入根接点 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) { //输出数据 System.out.print(elements[front].data + " "); //装入左树 if (elements[front].leftTree != null) { elements[rear++] = elements[front].leftTree; } if (elements[front].rightTree != null) { elements[rear++] = elements[front].rightTree; } //数组向前移动 front++; } } catch (Exception e) { break; } } } //求二叉树的高度 public static int height(BiTree tree) { if (tree == null) return 0; else {//递归饭会树的高度 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; //计算左树 leftLevel = leftTree == null ? -1 : leftTree.level(data); //计算左树 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(递归) 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;//增加一个叶子节点 else return left + right; } } //求二叉树父节点个数 public static int fatherNodes(BiTree tree) { //节点为空或者单节点 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; else return 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 || tree.leftTree==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; else return 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; else return 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); else return (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();//交换左树 } if (rightTree != null) { rightTree.exChange();//交换右树 } 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;//求和 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) { BiTree l = new BiTree(10); BiTree m = new BiTree(9); BiTree n = new BiTree(8); BiTree h = new BiTree(7); BiTree g = new BiTree(6); BiTree e = new BiTree(5,n,l); BiTree d = new BiTree(4,m,null); BiTree c = new BiTree(3,g,h); BiTree b = new BiTree(2, d, e); 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)); } } 小结:大家可以来练习一下，自己写一遍 熟悉一下. |
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