前两题写的又快又准 , 第三题写的又慢又乱

235. 二叉搜索树的最近公共祖先

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

701.二叉搜索树中的插入操作

public TreeNode insertIntoBST(TreeNode root, int val) {
        if (root == null) {
            return new TreeNode(val);
        }
        if (val < root.val) {
            root.left = insertIntoBST(root.left, val);
        } else {
            root.right = insertIntoBST(root.right, val);
        }
        return root;
    }

    public TreeNode insertIntoBST2(TreeNode root, int val) {
        if (root == null) {
            return new TreeNode(val);
        }
        TreeNode cur = root, pre = null;
        while (cur != null) {
            if (cur.val < val) {
                pre = cur;
                cur = cur.right;
            } else {
                pre = cur;
                cur = cur.left;
            }
        }
        if (pre.val < val) {
            pre.right = new TreeNode(val);
        } else {
            pre.left = new TreeNode(val);
        }
        return root;

    }

450.删除二叉搜索树中的节点

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 {
            if(root.left == null) return root.right;
            if(root.right == null) return root.left;
            TreeNode right = root.right;
            while(right.left != null){
                right = right.left;
            }
            root.val = right.val;
            root.right=deleteNode(root.right,right.val);
        }
        return root;
    }