Insert Node in a Binary Search Tree 85

Question

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.

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

Recursion: 若插入node值大于root值，则插入root右子树，若小于则插入root左子树

Non-Recursion: 用stack实现。若插入的node值大于root值，则将root加入stack，将root变为其右孩子，反之则变为其左孩子，直到root为null为止，此时stack的栈顶元素即为node要插入的元素，node值大于该元素则插在右边，否则插在左边。

代码如下：

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) {
        if (root == null) {
            return node;
        }
        if (root.val > node.val) {
            root.left = insertNode(root.left, node);
        } else {
            root.right = insertNode(root.right, node);
        }
        return root;
    }
}

Non-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) {
        if (root == null) {
            root = node;
            return root;
        }
        TreeNode tmp = root;
        TreeNode last = null;
        while (tmp != null) {
            last = tmp;
            if (tmp.val > node.val) {
                tmp = tmp.left;
            } else {
                tmp = tmp.right;
            }
        }
        if (last != null) {
            if (last.val > node.val) {
                last.left = node;
            } else {
                last.right = node;
            }
        }
        return root;
    }
}