Insert Node in a Binary Search Tree

Insert Node in a Binary Search Tree

Given a binary search tree and a new tree node, insert the node into the tree. You should keep the tree still be a valid binary search tree.

Example

Given binary search tree as follow, after Insert node 6, the tree should be:

  2             2

 / \           / \

1   4   -->   1   4

   /             / \ 

  3             3   6

Challenge

Can you do it without recursion?

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of the binary search tree.
     * @param node: insert this node into the binary search tree
     * @return: The root of the new binary search tree.
     */
    public TreeNode insertNode(TreeNode root, TreeNode node) {
        // write your code here
        if (root == null) {
            return node;
        }
        
        if (root.val > node.val) {
            root.left = insertNode(root.left, node);
        }
        if (root.val < node.val) {
            root.right = insertNode(root.right, node);
        }
        
        return root;
    }
}

no recursion
从上到下找到需要插入的位置，用cur记录插入点的parent

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of the binary search tree.
     * @param node: insert this node into the binary search tree
     * @return: The root of the new binary search tree.
     */
    public TreeNode insertNode(TreeNode root, TreeNode node) {
        // write your code here
        if (root == null) {
            return node;
        }
        
        TreeNode next = root;
        TreeNode cur = null;
        
        while (next != null) {
            cur = next;
            if (next.val > node.val) {
                next = next.left;
            }
            else {
                next = next.right;
            }
        }
        
        if (cur != null) {
            if (cur.val > node.val) {
                cur.left = node;
            }
            else {
                cur.right = node;
            }
        }
        
        return root;
    }
}