（二）3道簡單算法：二叉樹最大節點、十進制轉任意進制、位運算實現加法

    //【1】.**************************************************************************

    /**

      問題描述：前中後序，遞歸，查找二叉樹最大節點

     */
   /**
     二叉樹的最大節點
     在二叉樹中尋找值最大的節點並返回。
     */
    public class TreeNode {
        public int value;
        public TreeNode leftTreeNode;
        public TreeNode rightTreeNode;
        public TreeNode(int value, TreeNode leftTreeNode, TreeNode rightTreeNode) {
            this.value = value;
            this.leftTreeNode = leftTreeNode;
            this.rightTreeNode = rightTreeNode;
        }
    }

    private static TreeNode maxTreeNode;
    /**
     * 考點：二叉樹-深度遍歷-遞歸-前序遍歷
     * @param root
     * @return
     */
    public TreeNode maxNodePre(TreeNode root) {
        if (root == null) {
            return null;
        }
        maxTreeNode = root;
        nextTreeNodePre(root);
        return maxTreeNode;
    }

    private void nextTreeNodePre(TreeNode treeNode) {
        System.out.println(treeNode.value);

        if (treeNode.value > maxTreeNode.value) {
            maxTreeNode = treeNode;
        }
        if (treeNode.leftTreeNode != null) {
            nextTreeNodePre(treeNode.leftTreeNode);
        }

        if (treeNode.rightTreeNode != null) {
            nextTreeNodePre(treeNode.rightTreeNode);
        }
    }

    /**
     * 考點：二叉樹-深度遍歷-遞歸-中序遍歷
     * @param root
     * @return
     */
    public TreeNode maxNodeMid(TreeNode root) {
        if (root == null) {
            return null;
        }
        maxTreeNode = root;
        nextTreeNodeMid(root);
        return maxTreeNode;
    }

    private void nextTreeNodeMid(TreeNode treeNode) {
        if (treeNode.leftTreeNode != null) {
            nextTreeNodeMid(treeNode.leftTreeNode);
        }

        System.out.println(treeNode.value);

        if (treeNode.value > maxTreeNode.value) {
            maxTreeNode = treeNode;
        }

        if (treeNode.rightTreeNode != null) {
            nextTreeNodeMid(treeNode.rightTreeNode);
        }
    }

    /**
     * 考點：二叉樹-深度遍歷-遞歸-後序遍歷
     * @param root
     * @return
     */
    public TreeNode maxNodeAfter(TreeNode root) {
        if (root == null) {
            return null;
        }
        maxTreeNode = root;
        nextTreeNodeAfter(root);
        return maxTreeNode;
    }

    private void nextTreeNodeAfter(TreeNode treeNode) {
        if (treeNode.leftTreeNode != null) {
            nextTreeNodeAfter(treeNode.leftTreeNode);
        }

        if (treeNode.rightTreeNode != null) {
            nextTreeNodeAfter(treeNode.rightTreeNode);
        }

        System.out.println(treeNode.value);

        if (treeNode.value > maxTreeNode.value) {
            maxTreeNode = treeNode;
        }
    }

   //【2】*************************************************************************************
    /**
     * 問題描述：進制轉換
      給定一個十進制數 n 和 一個整數 k, 將 十進制數 n 轉換成 k進制數.
      【方法】:遞歸整除和餘數
     * @param n
     * @param k
     * @return
     */
    public String hexConversion(int n, int k) {
        StringBuffer resultNumber = new StringBuffer();
        tenToK(resultNumber, n, k);
        System.out.println("n:k:result: " + n +" "+ k + " " + resultNumber.toString());

        return resultNumber.toString();
    }

    private void tenToK(StringBuffer stringBuffer, int n, int k) {
        int integral = n/k;
        int mode = n % k;
        stringBuffer.insert(0, mode);
        if (integral >= k) {
            tenToK(stringBuffer, integral, k);
        } else if (integral > 0) {
            stringBuffer.insert(0, integral);
        }
    }

    private void testHexConversion() {
        hexConversion(3, 4);
        hexConversion(9, 4);
        hexConversion(29, 4);
        hexConversion(3, 5);
        hexConversion(9, 5);
        hexConversion(29, 5);
    }

   //【3】**************************************************************
   /**

    位運算實現加法

    */
    public int aplusb(int a, int b) {
        int sum_without_carry, carry;

        sum_without_carry = a^b; //沒有進位的和
        carry = (a&b)<<1; //進位
        if(carry==0)
            return sum_without_carry;
        else
            return aplusb(sum_without_carry, carry);
    }

    public void testAplusB() {
        int result = aplusb(4,19);
        System.out.println(result);
    }