[Lintcode] 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?

SOLUTION 1:

首先从最好理解的recursion开始。

递归算法，1，首先这个输入跟我要做递归的函数input相同，所以没必要再做一个辅助函数，2，然后，再看递归具体步骤，从局部考虑，如果node.val > root.val 递归应该向右边进行（也就是说node一定会加在root右子树上），同理node.val < root.val ==>递归向左进行。3，最后考虑递归的结束条件是什么，这题的结束条件就是指针移动到一个没有左儿子或者右儿子的地方，把node加进去，返回这个结果 ==> 也就是root.left/right = node。

具体看代码：

/**
 * 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) {
        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;
    }
}

SOLUTION 2:

非递归方法：

非递归就是说，找到root点，root左子树或者右子树是空的，选择性的把node放在上面。

/**
 * 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) {
        if (root == null){
            return node;
        }
        if (node == null){
            return root;
        }
        TreeNode current = root;
        while (current != null){
            if (node.val > current.val && current.right == null){
                current.right = node;
                break;
            } else if (node.val < current.val && current.left == null){
                current.left = node;
                break;
            } else if (node.val > current.val){
                current = current.right;
            } else {
                current = current.left;
            }
        }
        return root;
    }
}