给定一棵二叉查找树和一个新的树节点，将节点插入到树中。
你需要保证该树仍然是一棵二叉查找树。
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.

样例
给出如下一棵二叉查找树，在插入节点6之后这棵二叉查找树可以是这样的：
2 2
/ \ / \
1 4 –>1 4
/ / \
3 3 6

挑战
能否不使用递归？

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. */
    //递归
    public TreeNode insertNode(TreeNode root, TreeNode node) {
        if(root == null) {
            root = node;
            return root;
        } else if(root.val < node.val) {
            root.right = insertNode(root.right, node);
            return root;
        }else {
            root.left = insertNode(root.left,node);
            return root;
        }
    }

     //二叉查找树节点插入只插在叶子节点处
     //非递归
    public TreeNode insertNode(TreeNode root, TreeNode node) {
        if(root == null) {
            root = node;
            return root;
        }
        TreeNode p = root;
        while(p != null) {
            if(p.val < node.val) {
                if(p.right == null) {
                    p.right = node;
                    return root;
                }
                p = p.right;
            }else {
                if(p.left == null) {
                    p.left = node;
                    return root;
                }
                p = p.left;
            }
        }
        return root;
    }
}