java算法面试题之一

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题目：int values[] = {80099, 16114, 63108, 25032, 31044, 59069, 39099, 13110, 34101, 66120,19116, 72105, 70045, 38032, 41110, 12105, 75110, 27105, 1105, 9114,67117, 20101, 21100, 11032, 79046, 32112, 5111, 6117, 45116, 22032,61097, 65120, 49110, 15101, 82109, 50103, 54110, 17101, 46032, 4121,
76097, 7032, 57105, 2102, 58044, 8097, 44099, 73064, 81111, 43097,30112, 14116, 60109, 74104, 77105, 35097, 64058, 29112, 55032, 33108,71108, 40111, 47088, 52117, 56076, 68097, 37101, 78114, 24110, 53097,69110, 48105, 18115, 26072, 3032, 42116, 62105, 51120, 28065, 10101,23105, 36115}
这样一个数组，操作如下。
1.首先将数组的内容有序的存入二叉树中。
2.然后由小到大遍历二叉树。
3.把遍历到的数值取后三位转化成字符串输出。

思路：
- 首先要将数组的内容有序的存入二叉树中，那么我们可以先将数组排序，排序之后再插入到二叉树中，然后采用层级遍历二叉树的方法，输出数值，在输出数值的时候用String的方法截取后三位。
代码如下:

public class TreeNode {    public int value;    public TreeNode left;    public TreeNode right;    public TreeNode(int value) {        this(value, null, null);    }    public TreeNode(int value, TreeNode left, TreeNode right) {        this.value = value;        this.left = left;        this.right = right;    }}// 对二叉树的操作public class Tree {    public TreeNode root;    public Tree(TreeNode root) {        this.root = root;    }    /***     * 前序遍历     * @param node     */    public void preLeft(TreeNode node) {        if (node != null) {            System.out.print(node.value + "\t");            preLeft(node.left);            preLeft(node.right);        }    }    /***     * 中序遍历     * @param node     */    public void preMiddle(TreeNode node) {        if (node != null) {            preLeft(node.left);            System.out.print(node.value + "\t");            preLeft(node.right);        }    }    /**     * 后序遍历     * @param node     */    public void preRight(TreeNode node) {        if (node != null) {            preLeft(node.left);            preLeft(node.right);            System.out.print(node.value + "\t");        }    }    /***     * 层序遍历     * @param root     */    public static void preLevel(TreeNode root) {        if (root == null) return;        Queue<TreeNode> q = new LinkedList<TreeNode>();        q.add(root);        while (!q.isEmpty()) {            TreeNode temp = q.poll();            System.out.print(temp.value + "\t");            if (temp.left != null) q.add(temp.left);            if (temp.right != null) q.add(temp.right);        }    }    /***     * 层序遍历     * @param treeNode     */    public void preLevel2(TreeNode treeNode) {        if (treeNode != null) {            Queue<TreeNode> queue = new LinkedList<>();            queue.add(treeNode);            while (!queue.isEmpty()) {                TreeNode node = queue.poll();                System.out.print(convert(node.value) + "\t");                if (node.left != null) {                    queue.add(node.left);                }                if (node.right != null) {                    queue.add(node.right);                }            }        }    }    /**     * 拿到数值的后三位     * @param value     * @return     */    public String convert(int value){        String valueStr = String.valueOf(value);        int length = valueStr.length();        if(length > 3){            return valueStr.substring(length-3);        }        return null;    }    /***     * 获取所有结点     * @param node     * @return     */    public int getCountNode(TreeNode node) {        int count = 0;        if (node != null) {            count += 1;            count += getCountNode(node.left);            count += getCountNode(node.right);        }        return count;    }    /**     * 树的深度     *     * @param treeNode     * @return     */    public int getDepth(TreeNode treeNode) {        if (treeNode != null) {            int leftDepth = getDepth(treeNode.left);            int rightDepth = getDepth(treeNode.right);            return 1 + ((leftDepth > rightDepth) ? leftDepth : rightDepth);        }        return 0;    }}// 测试public class Test {    public static void main(String[] args) {        int values[] = {80099, 16114, 63108, 25032, 31044, 59069, 39099, 13110, 34101, 66120, 19116, 72105, 70045, 38032, 41110, 12105, 75110, 27105, 1105, 9114, 67117, 20101, 21100, 11032, 79046, 32112, 5111, 6117, 45116, 22032, 61097, 65120, 49110, 15101, 82109, 50103, 54110, 17101, 46032, 4121,                76097, 7032, 57105, 2102, 58044, 8097, 44099, 73064, 81111, 43097, 30112, 14116, 60109, 74104, 77105, 35097, 64058, 29112, 55032, 33108, 71108, 40111, 47088, 52117, 56076, 68097, 37101, 78114, 24110, 53097, 69110, 48105, 18115, 26072, 3032, 42116, 62105, 51120, 28065, 10101, 23105, 36115};        int length = values.length;        // 使用冒泡排序将数组有序化        for (int i = 0; i < length; i++) {            for (int j = 0; j < length - i - 1; j++) {                if (values[j] > values[j + 1]) {                    int temp = values[j];                    values[j] = values[j + 1];                    values[j + 1] = temp;                }            }        }        for (int i = 0; i < length; i++) {            System.out.print(values[i] + "\t");        }        // 生成一颗二叉树        TreeNode root = createBinaryTree(values, 0);        Tree tree = new Tree(root);        // 使用各种方法遍历        System.out.println("\n层序遍历");        tree.preLevel(root);        System.out.println("\n层序遍历打印后三位");        tree.preLevel2(root);        System.out.println("\n左序遍历");        tree.preLeft(root); // 左序遍历        System.out.println("\n中序遍历");        tree.preMiddle(root); // 中序遍历        System.out.println("\n右序遍历");        tree.preRight(root); // 右序遍历        System.out.println("\n树的深度是");        int count = tree.getDepth(root);        System.out.println(count);    }    /***     * 创建一颗二叉树     * @param values 值     */    public static TreeNode createBinaryTree(int[] values, int index) {        TreeNode root = null;        if (index < values.length) {            root = new TreeNode(values[index]);            root.left = createBinaryTree(values, 2 * index + 1);            root.right = createBinaryTree(values, 2 * index + 2);        }        return root;    }}

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