leetcode--CountCompleteTreeNodes

complete binary tree:最后一层以上的节点全部排满，最后一层叶子节点都在左边。（可能满也可能不满）

如果用前序中序或者后序递归的话，内存过大，时间也不够。因此根据complete binary tree的性质，分别求最左子节点和最右子节点的深度，如果相等就根据公式返回，不等再向下递归。这样可以减少递归的次数。

/**
 * Created by marsares on 15/6/16.
 */
public class CountCompleteTreeNodes {
    public int countNodes(TreeNode root) {
        if(countLeft(root,0)==countRight(root,0))return twoPower(countLeft(root,0))-1;
        else return countNodes(root.left)+countNodes(root.right)+1;
    }
    private int countLeft(TreeNode root,int height){
        if(root==null)return height;
        return countLeft(root.left,height+1);
    }
    private int countRight(TreeNode root,int height){
        if(root==null)return height;
        return countRight(root.right, height + 1);
    }
    private int twoPower(int n){
        int ans=1;
        for(int i=0;i<n;i++){
            ans=ans*2;
        }
        return ans;
    }
    public static void main(String[]args){
        CountCompleteTreeNodes cctn=new CountCompleteTreeNodes();
        BinaryTreeSerialize bts=new BinaryTreeSerialize();
        TreeNode root=bts.Unserialize("{5,3,8,2,4,7}");
        System.out.println(cctn.countNodes(root));
    }
}