Java实现二叉树的代码怎么写

本篇内容主要讲解“Java实现二叉树的代码怎么写”，感兴趣的朋友不妨来看看。本文介绍的方法操作简单快捷，实用性强。下面就让小编来带大家学习“Java实现二叉树的代码怎么写”吧!

以此图为例，完整代码如下：

//基础二叉树实现//使用左右孩子表示法 import java.util.*;import java.util.Deque; public class myBinTree {    private static class Treenode{        char val;        TreeNode left;        TreeNode right;         public TreeNode(char val) {            this.val = val;        }    }     public static TreeNode build(){        TreeNode nodeA=new TreeNode('A');        TreeNode nodeB=new TreeNode('B');        TreeNode nodeC=new TreeNode('C');        TreeNode nodeD=new TreeNode('D');        TreeNode nodeE=new TreeNode('E');        TreeNode nodeF=new TreeNode('F');        TreeNode nodeG=new TreeNode('G');        TreeNode nodeH=new TreeNode('H');        nodeA.left=nodeB;        nodeA.right=nodeC;        nodeB.left=nodeD;        nodeB.right=nodeE;        nodeE.right=nodeH;        nodeC.left=nodeF;        nodeC.right=nodeG;        return nodeA;    }     //方法1(递归)    //先序遍历: 根左右    public static void preOrder(TreeNode root){        if(root==null){            return;        }        System.out.print(root.val+" ");        preOrder(root.left);        preOrder(root.right);    }     //方法1(递归)    //中序遍历    public static void inOrder(TreeNode root){        if(root==null){            return;        }        inOrder(root.left);        System.out.print(root.val+" ");        inOrder(root.right);    }     //方法1(递归)    //后序遍历    public static void postOrder(TreeNode root){        if(root==null){            return;        }        postOrder(root.left);        postOrder(root.right);        System.out.print(root.val+" ");    }     //方法2(迭代)    //先序遍历 (迭代)    public static void preOrderNonRecursion(TreeNode root){        if(root==null){            return ;        }        Deque<TreeNode> stack=new LinkedList<>();        stack.push(root);        while (!stack.isEmpty()){            TreeNode cur=stack.pop();            System.out.print(cur.val+" ");            if(cur.right!=null){                stack.push(cur.right);            }            if(cur.left!=null){                stack.push(cur.left);            }        }    }     //方法2(迭代)    //中序遍历 (迭代)    public static void inorderTraversalNonRecursion(TreeNode root) {        if(root==null){            return ;        }         Deque<TreeNode> stack=new LinkedList<>();        // 当前走到的节点        TreeNode cur=root;        while (!stack.isEmpty() || cur!=null){            // 不管三七二十一，先一路向左走到根儿~            while (cur!=null){                stack.push(cur);                cur=cur.left;            }            // 此时cur为空，说明走到了null，此时栈顶就存放了左树为空的节点            cur=stack.pop();            System.out.print(cur.val+" ");            // 继续访问右子树            cur=cur.right;        }    }     //方法2(迭代)    //后序遍历 (迭代)    public static void postOrderNonRecursion(TreeNode root){        if(root==null){            return;        }        Deque<TreeNode> stack=new LinkedList<>();        TreeNode cur=root;        TreeNode prev=null;         while (!stack.isEmpty() || cur!=null){            while (cur!=null){                stack.push(cur);                cur=cur.left;            }             cur=stack.pop();            if(cur.right==null || prev==cur.right){                System.out.print(cur.val+" ");                prev=cur;                cur=null;            }else {                stack.push(cur);                cur=cur.right;            }        }    }     //方法1(递归)    //传入一颗二叉树的根节点，就能统计出当前二叉树中一共有多少个节点，返回节点数    //此时的访问就不再是输出节点值，而是计数器 + 1操作    public static int getNodes(TreeNode root){        if(root==null){            return 0;        }        return 1+getNodes(root.left)+getNodes(root.right);    }     //方法2(迭代)    //使用层序遍历来统计当前树中的节点个数    public static int getNodesNoRecursion(TreeNode root){        if(root==null){            return 0;        }        int size=0;        Deque<TreeNode> queue=new LinkedList<>();        queue.offer(root);        while (!queue.isEmpty()) {            TreeNode cur = queue.poll();            size++;            if (cur.left != null) {                queue.offer(cur.left);            }            if (cur.right != null) {                queue.offer(cur.right);            }        }        return size;    }     //方法1(递归)    //传入一颗二叉树的根节点，就能统计出当前二叉树的叶子结点个数    public static int getLeafNodes(TreeNode root){        if(root==null){            return 0;        }        if(root.left==null && root.right==null){            return 1;        }        return getLeafNodes(root.left)+getLeafNodes(root.right);    }     //方法2(迭代)    //使用层序遍历来统计叶子结点的个数    public static int getLeafNodesNoRecursion(TreeNode root){        if(root==null){            return 0;        }        int size=0;        Deque<TreeNode> queue=new LinkedList<>();        queue.offer(root);        while (!queue.isEmpty()){            TreeNode cur=queue.poll();            if(cur.left==null && cur.right==null){                size++;            }            if(cur.left!=null){                queue.offer(cur.left);            }            if(cur.right!=null){                queue.offer(cur.right);            }        }        return size;    }     //层序遍历    public static void levelOrder(TreeNode root) {        if(root==null){            return ;        }         // 借助队列来实现遍历过程        Deque<TreeNode> queue =new LinkedList<>();        queue.offer(root);        while (!queue.isEmpty()){            int size=queue.size();            for (int i = 0; i < size; i++) {                TreeNode cur=queue.poll();                System.out.print(cur.val+" ");                if(cur.left!=null){                    queue.offer(cur.left);                }                if(cur.right!=null){                    queue.offer(cur.right);                }            }        }    }     //传入一个以root为根节点的二叉树，就能求出该树的高度    public static int height(TreeNode root){        if(root==null){            return 0;        }        return 1+ Math.max(height(root.left),height(root.right));    }     //求出以root为根节点的二叉树第k层的节点个数    public static int getKLevelNodes(TreeNode root,int k){        if(root==null || k<=0){            return 0;        }        if(k==1){            return 1;        }        return getKLevelNodes(root.left,k-1)+getKLevelNodes(root.right,k-1);    }     //判断当前以root为根节点的二叉树中是否包含指定元素val，    //若存在返回true，不存在返回false    public static boolean contains(TreeNode root,char value){        if(root==null){            return false;        }        if(root.val==value){            return true;        }        return contains(root.left,value) || contains(root.right,value);    }      public static void main(String[] args) {        TreeNode root=build();         System.out.println("方法1(递归)：前序遍历的结果为:");        preOrder(root);        System.out.println();        System.out.println("方法2(迭代)：前序遍历的结果为:");        preOrderNonRecursion(root);        System.out.println();         System.out.println("方法1(递归)：中序遍历的结果为:");        inOrder(root);        System.out.println();        System.out.println("方法2(迭代)：中序遍历的结果为:");        inorderTraversalNonRecursion(root);        System.out.println();         System.out.println("方法1(递归)：后序遍历的结果为:");        postOrder(root);        System.out.println();        System.out.println("方法2(迭代)：后序遍历的结果为:");        postOrderNonRecursion(root);        System.out.println();        System.out.println();         System.out.println("层序遍历的结果为:");        levelOrder(root);        System.out.println();        System.out.println();         System.out.println("方法1(递归)：当前二叉树一共有:"+getNodes(root)+"个节点数");        System.out.println("方法2(迭代)：当前二叉树一共有:"+getNodesNoRecursion(root)+"个节点数");        System.out.println("方法1(递归)：当前二叉树一共有:"+getLeafNodes(root)+"个叶子节点数");        System.out.println("方法2(迭代)：当前二叉树一共有:"+getLeafNodesNoRecursion(root)+"个叶子节点数");        System.out.println(contains(root,'E'));        System.out.println(contains(root,'P'));        System.out.println("当前二叉树的高度为:"+height(root));        System.out.println("当前二叉树第3层的节点个数为:"+getKLevelNodes(root,3));    }}

如上main引用结果如下：

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