Java에서 이진 트리를 구현하는 코드를 작성하는 방법

이 사진을 예로 들어보겠습니다. 전체 코드는 다음과 같습니다.

//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,&#39;E&#39;));
        System.out.println(contains(root,&#39;P&#39;));
        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));
    }
}

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