(medium)LeetCode 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.

解法一：递归超时

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) {
        if(root==null)return 0;
        int num1=0,num2=0;
        num1=count(root.left);
        num2=count(root.right);
        return 1+num1+num2;  

    }
    int count(TreeNode root){
        if(root==null) return 0;
        int num1=0,num2=0;
        if(root.left==null && root.right==null)
            return 1;
        if(root.left!=null)
            num1=count(root.left);
        if(root.right!=null)
            num2=count(root.right);
         return 1+num1+num2;
    }
}

　　运行结果：

解法二：层次遍历，队列,超时

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) {
        if(root==null)return 0;

        Queue<TreeNode> q=new LinkedList<>();
        q.offer(root);
        int num=1;
        TreeNode tmp=null;
        while(!q.isEmpty()){
            tmp=q.poll();
            if(tmp.left!=null){
                q.offer(tmp.left);
                num++;
            }
            if(tmp.right!=null){
                q.offer(tmp.right);
                num++;
            }
        }
        return num;
    }

}

　　解法三：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)

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) {
        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;
    }
}

　　运行结果：