模拟实现二叉树的基本操作

import java.util.*;
class BinaryTree {
    static class TreeNode {
        public char val;
        public TreeNode left;//左孩子的引用
        public TreeNode right;//右孩子的引用
        public TreeNode(char val) {
            this.val = val;
        }
    }

    //创建一棵二叉树,返回这棵树的根节点
    public TreeNode createTree() {
        TreeNode A = new TreeNode('A');
        TreeNode B = new TreeNode('B');
        TreeNode C = new TreeNode('C');
        TreeNode D = new TreeNode('D');
        TreeNode E = new TreeNode('E');
        TreeNode F = new TreeNode('F');
        TreeNode G = new TreeNode('G');
        TreeNode H = new TreeNode('H');
        A.left = B;
        A.right = C;
        B.left = D;
        B.right = E;
        C.left = F;
        C.right = G;
        E.right = H;
        return A;
    }

    //前序遍历1
    public void preOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        System.out.print(root.val + " ");
        preOrder(root.left);
        preOrder(root.right);
    }
    //前序遍历2
    public List<Character> preorderTraversal(TreeNode root) {//如果<>里是TreeNode，则下面是list.add(root)
        List<Character> list = new ArrayList<>();//向上转型，每递归一次就创建一个List类型的变量list
        if (root == null) {
            return list;
        }
        list.add(root.val);
        List<Character> leftTree = preorderTraversal(root.left);
        list.addAll(leftTree);
        List<Character> rightTree = preorderTraversal(root.right);
        list.addAll(rightTree);
        return list;
    }

    //中序遍历1
    void inOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        inOrder(root.left);
        System.out.print(root.val + " ");
        inOrder(root.right);
    }
    //中序遍历2
    public List<Character> inorderTraversal(TreeNode root) {
        List<Character> list = new ArrayList<>();
        if (root == null) {
            return list;
        }
        List<Character> leftTree = inorderTraversal(root.left);
        list.addAll(leftTree);
        list.add(root.val);
        List<Character> rightTree = inorderTraversal(root.right);
        list.addAll(rightTree);
        return list;
    }

    //后序遍历1
    void postOrder(TreeNode root) {
        if (root == null) {
            return;
        }
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val + " ");
    }
    //后序遍历2
    public List<Character> postorderTraversal(TreeNode root) {
        List<Character> list = new ArrayList<>();
        if (root == null) {
            return list;
        }
        List<Character> leftTree = postorderTraversal(root.left);
        list.addAll(leftTree);
        List<Character> rightTree = postorderTraversal(root.right);
        list.addAll(rightTree);
        list.add(root.val);
        return list;
    }

    public static int nodeSize;
    //获取树中节点的个数：遍历思路
    void size(TreeNode root) {//返回值为void
        if (root == null) {
            return;
        }
        nodeSize++;
        size(root.left);
        size(root.right);
    }
    //获取树中节点的个数：子问题思路
    int size2(TreeNode root) {//返回值为int
        if (root == null) {
            return 0;
        }
        return size2(root.left) + size2(root.right) + 1;
    }

    public static int leafSize = 0;
    //获取叶子节点的个数：遍历思路
    void getLeafNodeCount1(TreeNode root) {
        if (root == null) {
            return;
        }
        if (root.left == null && root.right == null) {
            leafSize++;
        }
        getLeafNodeCount1(root.left);
        getLeafNodeCount1(root.right);
    }
    //获取叶子节点的个数：子问题思路
    int getLeafNodeCount2(TreeNode root) {
        if (root == null) {
            return 0;
        }
        if (root.left == null && root.right == null) {
            return 1;
        }
        return getLeafNodeCount2(root.left) + getLeafNodeCount2(root.right);
    }

    //获取第K层节点的个数
    int getKLevelNodeCount(TreeNode root, int k) {//root这棵树第k层节点数量=root.left这棵树第k-1层节点数量+root.right这棵树第k-1层节点数量
        if (root == null) {
            return 0;
        }
        if (k == 1) {//root不为null且k为1时，相当于属于第k层的一个节点数
            return 1;
        }
        return getKLevelNodeCount(root.left, k - 1) + getKLevelNodeCount(root.right, k - 1);
    }

    //获取二叉树的高度，时间复杂度：O(N)
    int getHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftTreeHeight = getHeight(root.left);
        int rightTreeHeight = getHeight(root.right);
        return leftTreeHeight > rightTreeHeight ? leftTreeHeight + 1 : rightTreeHeight + 1;
    }

    // 检测值为value的元素是否存在
    boolean find(TreeNode root, char key) {
        if(root == null) {
            return false;
        }
        if(root.val == key) {
            return true;
        }
        boolean leftVal = find(root.left, key);
        if(leftVal == true) {
            return true;
        }
        boolean rightVal = find(root.right, key);
        if(rightVal == true) {
            return true;
        }
        return false;
    }

    //层序遍历
    //法一：返回值为void
    void levelOrder1(TreeNode root) {
        if(root == null) {
            return;
        }
        Queue<TreeNode> queue = new LinkedList<>();//链表实现队列
        queue.offer(root);
        while(!queue.isEmpty()) {
            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);
            }
        }
    }
    //法二：返回值为List>
    List<List<TreeNode>> levelOrder2(TreeNode root) {
        List<List<TreeNode>> ret = new ArrayList<>();//顺序表里面的每个元素又是一个顺序表
        if(root == null) {
            return ret;
        }
        Queue<TreeNode> queue = new LinkedList<>();//链表实现队列
        queue.offer(root);
        while(!queue.isEmpty()) {
            int size = queue.size();//计算某一层一共有多少个节点
            List<TreeNode> tmp = new ArrayList<>();
            while(size != 0) {
                TreeNode cur = queue.poll();
                tmp.add(cur);
                size--;
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
            }
            ret.add(tmp);
        }
        return ret;
    }

    // 判断一棵树是不是完全二叉树
    boolean isCompleteTree(TreeNode root) {
        if (root == null) {
            return true;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            if(cur != null) {
                queue.offer(cur.left);
                queue.offer(cur.right);
            }else {//此时cur为null，即队列弹出第一个null
                break;
            }
        }
        while (!queue.isEmpty()) {//队列弹出第一个null后，若后面还能弹出非null的节点，则不是完全二叉树
            TreeNode cur = queue.poll();
            if (cur != null) {
                return false;
            }
        }
        return true;
    }
}