222. Count Complete Tree Nodes

Q:
Given a complete binary tree, count the number of nodes.
Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2^h nodes inclusive at the last level h.
 
A:
-------****** 第一遍-----递归---------------

public class Solution {
    public int countNodes(TreeNode root) {
        if(root == null)
            return 0;
        int l = getDepth(root.left);
        int r = getDepth(root.right);
        if(l==r)
            return 1<<(l+1)+countNodes(root.right);
        else
            return 1<<(r+1)+countNodes(root.left);
    }
    private int getDepth(TreeNode root){
        if(root == null)
            return 0;
        return 1+ getDepth(root.left);
    }
}

但上面的会LTE，   原因在于depth计算了多次。
试着把Recursive的 getDepth改成 iterative的 。 依然不行。


----------------------Solution 2 ----------------------
如果从某节点一直向左的高度 = 一直向右的高度, 那么以该节点为root的子树一定是complete binary tree. 而 complete binary tree的节点数,可以用公式算出 2^h - 1. 如果高度不相等, 则递归调用 return countNode(left) + countNode(right) + 1.  复杂度为O(h^2)  

 public class Solution {  
    public int countNodes(TreeNode root) {  
        if(root==null) return 0;  
          
        int l = getLeft(root) + 1;  
        int r = getRight(root) + 1;  
          
        if(l==r) {  
            return (2<<(l-1)) - 1;  
        } else {  
            return countNodes(root.left) + countNodes(root.right) + 1;  
        }  
    }  
      
    private int getLeft(TreeNode root) {  
        int count = 0;  
        while(root.left!=null) {  
            root = root.left;  
            ++count;  
        }  
        return count;  
    }  
      
    private int getRight(TreeNode root) {  
        int count = 0;  
        while(root.right!=null) {  
            root = root.right;  
            ++count;  
        }  
        return count;  
    }  
}  

***********上面的是每次查询了最左边和最右边的高度************

********3rd pass,   每次对比左右子树的高度，*********

public class Solution {
    public int countNodes(TreeNode root) {
        return helper(root,getHeight(root));
    }
    private int helper(TreeNode root, int height){
        if(root == null)
            return 0;
        int rightHeight = getHeight(root.right);
        if(height == rightHeight+1 ){
            return (1<<rightHeight) + helper(root.right, rightHeight);
        }else{ // height == rightHeight + 2
            return  helper(root.left,height-1) + (1<<rightHeight) ;
        }
    }
    private int getHeight(TreeNode root){
        int res = 0;
        while(root!=null){
            res++;
            root = root.left;
        }
        return res;
    }
}

Mistakes：
需要用 移位运算， 如果用Math.pow(a,b)的话，会超时