#85 Insert Node in a Binary Search Tree

题目描述：

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.

 Notice

You can assume there is no duplicate values in this tree + node.

Have you met this question in a real interview? 

Yes

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?

题目思路：

这题不用recursion做的话，就还是用BFS的思路。因为是BST，所以可以有方向的往下（要么左子树，要么右子树）。当遇到null的时候，就把新node连上去。

Mycode（AC = 42ms）：

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /**     * @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.     */    TreeNode* insertNode(TreeNode* root, TreeNode* node) {        // write your code here        if (root == NULL) return node;                queue<TreeNode *> helper;        helper.push(root);                while (!helper.empty()) {            TreeNode *tmp = helper.front();            helper.pop();                        if (node->val > tmp->val) {                if (tmp->right) {                    helper.push(tmp->right);                }                else {                    tmp->right = node;                }            }            else {                if (tmp->left) {                    helper.push(tmp->left);                }                else {                    tmp->left = node;                }            }        }                return root;    }};