一、 235. 二叉搜索树的最近公共祖先

题目连接：https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-search-tree/
思路：二叉搜索树最近的公共祖先的是第一次遇到root.val > p && root.val < q 或者 root.val< p && root.val > p区间
1、递归法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) return null;
        if (root.val > p.val && root.val > q.val) {
            return lowestCommonAncestor(root.left, p, q);
        } else if (root.val < p.val && root.val < q.val) {
            return lowestCommonAncestor(root.right, p, q);
        } else return root;
    }
}

2、迭代法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) return null;
        TreeNode cur = root;
        while (cur != null) {
            if (cur.val > p.val && cur.val > q.val) {
                cur = cur.left;
            } else if (cur.val < p.val && cur.val < q.val) {
                cur = cur.right;
            } else {
                return cur;
            }
        }
        return null;
    }
}

二、 701. 二叉搜索树中的插入操作

题目连接：https://leetcode.cn/problems/insert-into-a-binary-search-tree/
思路一：递归法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode insertIntoBST(TreeNode root, int val) {
        if (root == null) return new TreeNode(val);
        if (root.val > val) {
            root.left = insertIntoBST(root.left, val);
        } else {
            root.right = insertIntoBST(root.right, val);
        }
        return root;
    }
}

思路二：迭代法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode insertIntoBST(TreeNode root, int val) {
        TreeNode cur = root;
        while (cur != null) {
            if (cur.val > val) {
                if (cur.left == null) {
                    cur.left = new TreeNode(val);
                    return root;
                } else {
                    cur = cur.left;
                }
            } else {
                if (cur.right == null) {
                    cur.right = new TreeNode(val);
                    return root;
                } else {
                    cur = cur.right;
                }
            }
        }
        return new TreeNode(val);
    }
}

思路三：迭代法，记录前一个节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode insertIntoBST(TreeNode root, int val) {
        if (root == null) return new TreeNode(val);
        TreeNode cur = root;
        TreeNode pre = root;
        while (cur != null) {
            pre = cur;
            if (cur.val > val) {
                cur = cur.left;
            } else {
                cur = cur.right;
            }
        }
        if (pre.val > val) {
            pre.left = new TreeNode(val);
        } else {
            pre.right = new TreeNode(val);
        }
        return root;
    }
}

三、 450. 删除二叉搜索树中的节点

题目连接：https://leetcode.cn/problems/delete-node-in-a-bst/
思路：当root.val == key 时候，寻找root.right 是否为空，为空则返回root.left,不为空寻找root.right最小的节点，即TreeNode leftNode = root.right;
while (leftNode.left != null) {
leftNode = leftNode.left;
}
leftNode.left = root.left;
return root.right;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) return null;
        if (root.val > key) {
            root.left = deleteNode(root.left, key);
        } else if (root.val < key) {
            root.right = deleteNode(root.right, key);
        } else {
            TreeNode rightNode = root.right;
            if (rightNode == null) return root.left;
            TreeNode leftNode = root.right;
            while (leftNode.left != null) {
                leftNode = leftNode.left;
            }
            leftNode.left = root.left;
            return rightNode;
        }
        return root;
    }
}