二叉树的定义:
二叉树是树形结构的一个重要类型。许多实际问题抽象出来的数据结构往往是二叉树的形式,即使是一般的树也能简单地转换为二叉树,而且二叉树的存储结构及其算法都较为简单,因此二叉树显得特别重要。
??? 二叉树(BinaryTree)是n(n≥0)个结点的有限集,它或者是空集(n=0),或者由一个根结点及两棵互不相交的、分别称作这个根的左子树和右子树的二叉树组成。
??? 这个定义是递归的。由于左、右子树也是二叉树, 因此子树也可为空树。下图中展现了五种不同基本形态的二叉树。
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??? 其中 (a) 为空树, (b) 为仅有一个结点的二叉树, (c) 是仅有左子树而右子树为空的二叉树, (d) 是仅有右子树而左子树为空的二叉树, (e) 是左、右子树均非空的二叉树。这里应特别注意的是,二叉树的左子树和右子树是严格区分并且不能随意颠倒的,图 (c) 与图 (d) 就是两棵不同的二叉树。
二叉树的遍历
对于二叉树来讲最主要、最基本的运算是遍历。
??? 遍历二叉树 是指以一定的次序访问二叉树中的每个结点。所谓 访问结点 是指对结点进行各种操作的简称。例如,查询结点数据域的内容,或输出它的值,或找出结点位置,或是执行对结点的其他操作。遍历二叉树的过程实质是把二叉树的结点进行线性排列的过程。假设遍历二叉树时访问结点的操作就是输出结点数据域的值,那么遍历的结果得到一个线性序列。
从二叉树的递归定义可知,一棵非空的二叉树由根结点及左、右子树这三个基本部分组成。因此,在任一给定结点上,可以按某种次序执行三个操作:
??? (1)访问结点本身(N),
??? (2)遍历该结点的左子树(L),
??? (3)遍历该结点的右子树(R)。
以上三种操作有六种执行次序:
??? NLR、LNR、LRN、NRL、RNL、RLN。
注意:
????前三种次序与后三种次序对称,故只讨论先左后右的前三种次序。
由于被访问的结点必是某子树的根,所以N(Node)、L(Left subtlee)和R(Right subtree)又可解释为根、根的左子树和根的右子树。NLR、LNR和LRN分别又称为先根遍历、中根遍历和后根遍历。
二叉树的java实现
首先创建一棵二叉树如下图,然后对这颗二叉树进行遍历操作(遍历操作的实现分为递归实现和非递归实现),同时还提供一些方法如获取双亲结点、获取左孩子、右孩子等。

java实现代码:
- <span?abp="506"?style="font-size:14px;">package?study_02.datastructure.tree;??
- ??
- import?java.util.Stack;??
- ?
- ?
- ?*?@author?WWX?
- ?*/??
- public?class?BinaryTree?{??
- ??????
- ????private?TreeNode?root=null;??
- public?BinaryTree(){??
- ????????root=new?TreeNode(1,"rootNode(A)");??
- ????}??
- ?????
- ?????*?创建一棵二叉树?
- ?????*?<pre>?
- ?????*???????????A?
- ?????*?????B??????????C?
- ?????*??D?????E????????????F?
- ?????*??</pre>?
- ?????*?@param?root?
- ?????*?@author?WWX?
- ?????*/??
- void?createBinTree(TreeNode?root){??
- ????????TreeNode?newNodeB?=?2,"B");??
- ????????TreeNode?newNodeC?=?3,"C");??
- ????????TreeNode?newNodeD?=?4,"D");??
- ????????TreeNode?newNodeE?=?5,"E");??
- ????????TreeNode?newNodeF?=?6,"F");??
- ????????root.leftChild=newNodeB;??
- ????????root.rightChild=newNodeC;??
- ????????root.leftChild.leftChild=newNodeD;??
- ????????root.leftChild.rightChild=newNodeE;??
- ????????root.rightChild.rightChild=newNodeF;??
- ????}??
- ??????
- boolean?isEmpty(){??
- ????????return?root==null;??
- ??
- //树的高度??
- ????int?height(){??
- ????????return?height(root);??
- ??????
- int?size(){??
- return?size(root);??
- private?int?height(TreeNode?subTree){??
- if(subTree==null)??
- ????????????return?0;??
- else{??
- int?i=height(subTree.leftChild);??
- ????????????int?j=height(subTree.rightChild);??
- return?(i<j)?(j+1):(i+1);??
- ????????}??
- int?size(TreeNode?subTree){??
- null){??
- 0;??
- ????????}1+size(subTree.leftChild)??
- ????????????????????+size(subTree.rightChild);??
- ????????}??
- //返回双亲结点??
- public?TreeNode?parent(TreeNode?element){??
- return?(root==null||?root==element)?null:parent(root,?element);??
- public?TreeNode?parent(TreeNode?subTree,TreeNode?element){??
- return?if(subTree.leftChild==element||subTree.rightChild==element)??
- ??????????????
- return?subTree;??
- ????????TreeNode?p;??
- ??????????
- if((p=parent(subTree.leftChild,?element))!=null)??
- ??????????????
- return?p;??
- else??
- //递归在右子树中搜索??
- return?parent(subTree.rightChild,153); background-color:inherit; font-weight:bold">public?TreeNode?getLeftChildNode(TreeNode?element){??
- return?(element!=null)?element.leftChild:public?TreeNode?getRightChildNode(TreeNode?element){??
- null)?element.rightChild:public?TreeNode?getRoot(){??
- return?root;??
- //在释放某个结点时,该结点的左右子树都已经释放,??
- //所以应该采用后续遍历,当访问某个结点时将该结点的存储空间释放??
- void?destroy(TreeNode?subTree){??
- //删除根为subTree的子树??
- if(subTree!=null){??
- //删除左子树??
- ????????????destroy(subTree.leftChild);??
- //删除右子树??
- ????????????destroy(subTree.rightChild);??
- //删除根结点??
- ????????????subTree=void?traverse(TreeNode?subTree){??
- ????????System.out.println("key:"+subTree.key+"--name:"+subTree.data);;??
- ????????traverse(subTree.leftChild);??
- ????????traverse(subTree.rightChild);??
- //前序遍历??
- void?preOrder(TreeNode?subTree){??
- ????????????visted(subTree);??
- ????????????preOrder(subTree.leftChild);??
- ????????????preOrder(subTree.rightChild);??
- //中序遍历??
- void?inOrder(TreeNode?subTree){??
- ????????????inOrder(subTree.leftChild);??
- ????????????inOrder(subTree.rightChild);??
- //后续遍历??
- void?postOrder(TreeNode?subTree)?{??
- if?(subTree?!=?null)?{??
- ????????????postOrder(subTree.leftChild);??
- ????????????postOrder(subTree.rightChild);??
- //前序遍历的非递归实现??
- void?nonRecPreOrder(TreeNode?p){??
- ????????Stack<TreeNode>?stack=new?Stack<TreeNode>();??
- ????????TreeNode?node=p;??
- while(node!=null||stack.size()>0){??
- ????????????????visted(node);??
- ????????????????stack.push(node);??
- ????????????????node=node.leftChild;??
- ????????????}??
- ????????????<span?abp="507"?style="font-size:14px;">while</span>(stack.size()> ????????????????node=stack.pop();??
- ????????????????node=node.rightChild;??
- ????????????}???
- //中序遍历的非递归实现??
- void?nonRecInOrder(TreeNode?p){??
- ????????Stack<TreeNode>?stack?=new?Stack<BinaryTree.TreeNode>();??
- ????????TreeNode?node?=p;??
- 0){??
- //存在左子树??
- ????????????????stack.push(node);??
- ????????????????node=node.leftChild;??
- ????????????}??
- //栈非空??
- if(stack.size()> ????????????????node=node.rightChild;??
- //后序遍历的非递归实现??
- void?noRecPostOrder(TreeNode?p){??
- new?Stack<BinaryTree.TreeNode>();??
- ????????TreeNode?node?=p;??
- while(p!=//左子树入栈??
- for(;p.leftChild!=null;p=p.leftChild){??
- ????????????????stack.push(p);??
- //当前结点无右子树或右子树已经输出??
- null&&(p.rightChild==null||p.rightChild==node)){??
- ????????????????visted(p);??
- ??????????????????
- ????????????????node?=p;??
- ????????????????if(stack.empty())??
- ????????????????????return;??
- ????????????????p=stack.pop();??
- //处理右子树??
- ????????????stack.push(p);??
- ????????????p=p.rightChild;??
- void?visted(TreeNode?subTree){??
- ????????subTree.isVisted=true;??
- ????????System.out.println("key:"+subTree.key+"--name:"+subTree.data);;??
- ?????*?二叉树的节点数据结构?
- ?????*?@author?WWX?
- ?????*/??
- class??TreeNode{??
- int?key=0;??
- private?String?data=boolean?isVisted=false;??
- private?TreeNode?leftChild=private?TreeNode?rightChild= ??????????
- public?TreeNode(){}??
- ?????????*?@param?key??层序编码?
- ?????????*?@param?data?数据域?
- ?????????*/??
- public?TreeNode(int?key,String?data){??
- this.key=key;??
- this.data=data;??
- this.leftChild=this.rightChild=//测试??
- static?void?main(String[]?args)?{??
- ????????BinaryTree?bt?=?new?BinaryTree();??
- ????????bt.createBinTree(bt.root);??
- ????????System.out.println("the?size?of?the?tree?is?"?+?bt.size());??
- ????????System.out.println("the?height?of?the?tree?is?"?+?bt.height());??
- ????????System.out.println("*******(前序遍历)[ABDECF]遍历*****************");??
- ????????bt.preOrder(bt.root);??
- ??????????
- ????????System.out.println("*******(中序遍历)[DBEACF]遍历*****************");??
- ????????bt.inOrder(bt.root);??
- ?????????
- ????????System.out.println("*******(后序遍历)[DEBFCA]遍历*****************");??
- ????????bt.postOrder(bt.root);??
- ????????System.out.println("***非递归实现****(前序遍历)[ABDECF]遍历*****************");??
- ????????bt.nonRecPreOrder(bt.root);??
- ????????System.out.println("***非递归实现****(中序遍历)[DBEACF]遍历*****************");??
- ????????bt.nonRecInOrder(bt.root);??
- ????????System.out.println("***非递归实现****(后序遍历)[DEBFCA]遍历*****************");??
- ????????bt.noRecPostOrder(bt.root);??
- }??
- </span>??
?
输出结果
the size of the tree is 6
the height of the tree is 3
*******(前序遍历)[ABDECF]遍历*****************
key:1--name:rootNode(A)
key:2--name:B
key:4--name:D
key:5--name:E
key:3--name:C
key:6--name:F
*******(中序遍历)[DBEACF]遍历*****************
key:4--name:D
key:2--name:B
key:5--name:E
key:1--name:rootNode(A)
key:3--name:C
key:6--name:F
*******(后序遍历)[DEBFCA]遍历*****************
key:4--name:D
key:5--name:E
key:2--name:B
key:6--name:F
key:3--name:C
key:1--name:rootNode(A)
***非递归实现****(前序遍历)[ABDECF]遍历*****************
key:1--name:rootNode(A)
key:2--name:B
key:4--name:D
key:5--name:E
key:3--name:C
key:6--name:F
***非递归实现****(中序遍历)[DBEACF]遍历*****************
key:4--name:D
key:2--name:B
key:5--name:E
key:1--name:rootNode(A)
key:3--name:C
key:6--name:F
***非递归实现****(后序遍历)[DEBFCA]遍历*****************
key:4--name:D
key:5--name:E
key:2--name:B
key:6--name:F
key:3--name:C
key:1--name:rootNode(A)
來源:http://blog.csdn.net/wuwenxiang91322/article/details/12231657