public static Node insertBST(Node root, Node keyNode) {if (root== null) { //如果遇到空的地方，就是该插入的地方...root= keyNode;} else { //如果节点不为空，则需要查找对应的位置，和二分法查找类似if (root.data == keyNode.data) { //找到一个同值得数，直接返回return root;} else if (keyNode.data < root.data) {root.left = insertBST(root.left, keyNode);} else {root.right = insertBST(root.right, keyNode);}}return root;}

public static Node createBST(int[] arr) {Node root = null;for (int i = 0; i < arr.length; i++) {root = insertBST(root, new Node(arr[i]));}return root;}

public static boolean judgeBst(Node root) {int preVal = 0;Node cur = root;Stack<Node> stack = new Stack<>();while (true) {while (cur != null) {stack.push(cur);cur = cur.left;}if (stack.isEmpty())break;cur = stack.pop();if (preVal > cur.data) {return false;} else {preVal = cur.data;}cur = cur.right;}return true;}

public static Node deleteNodeFromBST(Node root, int key) {if (root == null) return null;if (key < root.data) {root.left = deleteNodeFromBST(root.left, key);} else if (key > root.data) {root.right = deleteNodeFromBST(root.right, key);} else {//找到keyif (root.left == null) {return root.right;} else if (root.right == null) {return root.left;}//有左右子树的情况...Node tmp = root;//找到root左子树中最右边的那个Node, 这里只能从tmp去遍历root = findRightest(tmp.left);root.left = deleteRightest(tmp.left);root.right = tmp.right;}return root;}// 找到左子树中最右的那个Node，因为它是root左子树的最大值...private static Node findRightest(Node root) {if (root == null) return null;while (root.right != null) {root = root.right;}return root;}// 删除刚刚从发现的那个点(root左子树中最右边的那个rightest)// 因为是最右面的Node，则必然没有右子树private static Node deleteRightest(Node root) {if (root.right == null)return root.left;root.right = deleteRightest(root.right);return null;}