给定二叉搜索树（BST）的根节点和要插入树中的值，将值插入二叉搜索树。 返回插入后二叉搜索树的根节点。 输入数据 保证 ，新值和原始二叉搜索树中的任意节点值都不同。

注意，可能存在多种有效的插入方式，只要树在插入后仍保持为二叉搜索树即可。 你可以返回 任意有效的结果 。

示例 1：

输入：root = [4,2,7,1,3], val = 5
输出：[4,2,7,1,3,5]
解释：另一个满足题目要求可以通过的树是：

示例 2：

输入：root = [40,20,60,10,30,50,70], val = 25
输出：[40,20,60,10,30,50,70,null,null,25]

示例 3：

输入：root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
输出：[4,2,7,1,3,5]

提示：

• 给定的树上的节点数介于 0 和 10^4 之间
• 每个节点都有一个唯一整数值，取值范围从 0 到 10^8
• -10^8 <= val <= 10^8
• 新值和原始二叉搜索树中的任意节点值都不同

Solution

• 递归法
• 找 ——> 改

# @lc code=start
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
        if not root:
            return TreeNode(val)
        if root.val==val:
            return root
        if root.val < val:
            root.right = self.insertIntoBST(root.right, val)
        if root.val > val:
            root.left = self.insertIntoBST(root.left, val)
        return root
# @lc code=end

cpp

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if (root == nullptr) {
            return root = new TreeNode(val);
        }
        if (root->val > val) {
            root->left = insertIntoBST(root->left, val);
        } else if (root->val < val) {
            root->right = insertIntoBST(root->right, val);
        }
        return root;
    }
};
// @lc code=end

• 迭代法
• 方法一：cur 搜索到空节点（待插入位置），参考 代码随想录

class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if (root == nullptr) {
            return root = new TreeNode(val);
        }

        TreeNode* cur = root, *pre = nullptr;
        while (cur != nullptr) {
            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;
    }
};

• 方法二：cur 搜索到待插入的结点（插入到该节点左/右孩子）

class Solution {
public:
    TreeNode* insertIntoBST(TreeNode* root, int val) {
        if (root == nullptr) {
            return root = new TreeNode(val);
        }

        TreeNode* cur = root;
        bool flag = 1;
        while (flag) {
            while (cur->val < val) {
                if (cur->right) {
                    cur = cur->right;
                } else {
                    cur->right = new TreeNode(val);
                    flag = 0;
                    break;
                }
            }
            while (cur->val > val) {
                if (cur->left) {
                    cur = cur->left;
                } else {
                    cur->left = new TreeNode(val);
                    flag = 0;
                    break;
                }
            }
        }
        return root;
    }
};

java

• 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) {
            TreeNode node = new TreeNode(val);
            return node;
        }
        if (root.val > val) {
            root.left = insertIntoBST(root.left, val);
        }
        if (root.val < val) {
            root.right = insertIntoBST(root.right, val);
        }
        return root;
    }
}

• 迭代法

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