701. 二叉搜索树中的插入操作

leetcode题目地址

利用二叉搜索树性质插入。

时间复杂度： $O(H)$
空间复杂度： $O(n)$

递归

// java
/**
 * 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 void insert(TreeNode root, int val){
        if(root == null) return;
        if(val > root.val && root.right == null){
            root.right = new TreeNode(val);
            return;
        }
        if(val < root.val && root.left == null){
            root.left = new TreeNode(val);
            return;
        }
        if(val > root.val) insert(root.right, val);
        if(val < root.val) insert(root.left, val);
    }

    public TreeNode insertIntoBST(TreeNode root, int val) {
        if(root == null) return new TreeNode(val);
        insert(root, val);
        return root;
    }
}

迭代一

// java
/**
 * 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 parent = root;
        while(cur != null){
            parent = cur;
            if(cur.val > val) cur = cur.left;
            else cur = cur.right;
        }
        cur = new TreeNode(val);
        if(val > parent.val) parent.right = cur;
        else parent.left = cur;
        return root;
    }
}

迭代二

// java
/**
 * 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;
        while(true){
            if(cur.val > val) {
                if(cur.left == null) {
                    cur.left = new TreeNode(val);
                    break;
                }
                cur = cur.left;
            } else {
                if(cur.right == null){
                    cur.right = new TreeNode(val);
                    break;
                }
                cur = cur.right;
            }
        }
        return root;
    }
}