Complete Binary Tree Insertion

Question: Given a complete binary tree, after inserting a node, it is still a complete binary tree.
Analysis: Use the properties of the complete binary tree to find the position to be inserted, first determine the height of the rightmost node of the left subtree and the leftmost node of the right subtree, if they are equal, only need to insert into the left subtree, otherwise insert right subtree.

accomplish:

/**题目：给完全二叉树插入一个节点，插完后保证还是完全二叉树
 * @author ShaoCheng
 * @version 1.0 2015-9-24
 */

class TreeNode{
    int val;
    TreeNode left, right;
    TreeNode(int key){
        val = key;
    }
}

public class Main {
    public static void main(String[] args){
        TreeNode p1 = new TreeNode(1);
        TreeNode p2 = new TreeNode(2);
        TreeNode p3 = new TreeNode(3);
        TreeNode p4 = new TreeNode(4);
        p1.left = p2;
        p1.right = p3;
        p2.left = p4;

        TreeNode node = new TreeNode(100);
        TreeNode root = insertBinary(p1, node);
        printTree(root);
    }

    //中序遍历
    public static void printTree(TreeNode root){
        if(root == null)
            return;
        System.out.println(root.val);
        printTree(root.left);
        printTree(root.right);
    }

    public static TreeNode insertBinary(TreeNode root, TreeNode node){
        if(root == null)
            return node;
        else if(node == null)
            return root;
        if(root.left == null){
            root.left = node;
            return root;
        }
        else if(root.right == null){
            root.right = node;
            return root;
        }

        int lrh = 0;
        int rrh = 0;
        TreeNode pNode = root.left;
        while(pNode != null){
            lrh++;
            pNode = pNode.right;
        }

        pNode = root.right;
        while(pNode != null){
            rrh++;
            pNode = pNode.left;
        }

        if(lrh != rrh){
            insertBinary(root.right, node);
        }
        else
            insertBinary(root.left, node);
        return root;
    }
}