222. Count Complete Tree Nodes

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 2h nodes inclusive at the last level h.

本题要求完全二叉树的节点个数，这里我们可以采用二分搜索的方法来做，二叉树的特点是每个非叶子节点都可以分成三部分，一部分是本身节点，一个是左子树，一个是右子树，我们先做一个方法函数来求一个树的高度，然后分别求出该树的高度和其右子树的高度（如果右子树非空），如果右子树的高度等于该树的高度-1，则说明该树的左子树部分是满的，则可以将其左子树的节点树和+递归其右子树；代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int height(TreeNode root){
        return root==null?-1:1+height(root.left);
    }
    public int countNodes(TreeNode root) {
        int h = height(root);
        return h<0?0:height(root.right)==h-1?(1<<h)+countNodes(root.right):(1<<h-1)+countNodes(root.left);
    }
}
// the run time complexity could be O(loglogn) this is because evey recision will take a height function ,will take longn time,and we have logn steps so the totl run time complexity could be loglogn;the space compexity could be logn; 

下面采用的是迭代的方法，代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int height(TreeNode node){
        return node==null?-1:1+height(node.left);
    }
    public int countNodes(TreeNode root) {
        int nodes = 0;
        int h = height(root);
        while(root!=null){
            if(height(root.right)==h-1){
                nodes+=(1<<h);
                root = root.right;
                h = height(root);
            }else{
                nodes+=(1<<h-1);
                root = root.left;
                h = height(root);
            }
        }
        return nodes;
    }
}

还有一种做法就是，创建两个TreeNode，left和right，当right不为空的时候，左子树一直遍历其左孩子，右子树一直遍历其右孩子，当right为空的时候，左面也为空，则说明没有下一行并且该行是满的；如果左面不为空，则递归其root的左子树和右子树；代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int countNodes(TreeNode root) {
        int height = 0;
        TreeNode left = root;
        TreeNode right = root;
        while(right!=null){
            height++;
            left = left.left;
            right = right.right;
        }
        if(left==null) return (1<<height)-1;
        return 1+countNodes(root.left)+countNodes(root.right);
    }
}