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.

满二叉树的节点数量的求解，直接遍历必然超时，所里判断是否是满二叉子树来做。

直接深度优先遍历DFS会超时，所以可以使用公式来计算，这个利用了满二叉树的性质

代码如下：

 /*class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } }*/ /* * 递归搜索的话必然超时 * 所以添加一下判断 * */ public class Solution { public int countNodes(TreeNode root) { if(root==null) return 0; int leftDepth = getLeft(root); int rightDepth = getRight(root); if(leftDepth == rightDepth) return (2<<leftDepth-1) - 1; else return countNodes(root.left) + countNodes(root.right) + 1; } public int getLeft(TreeNode root) { int count = 0; while(root!=null) { root = root.left; ++count; } return count; } public int getRight(TreeNode root) { int count = 0; while(root!=null) { root = root.right; ++count; } return count; } }

下面是C++的做法，就是做一个DFS深度优先搜索找到满二叉树，然后使用公式直接计算，否则递归计算

注意位运算的计算符号的优先级是低于普通的加减乘除运算的

代码如下：

#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>

using namespace std;

/* struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; */

class Solution 
{
public:
    int countNodes(TreeNode* root)
    {
        if (root == NULL)
            return 0;
        else
        {
            int leftDepth = getLeftDepth(root);
            int rightDepth = getRightDepth(root);
            if (leftDepth == rightDepth)
                return (2 << (leftDepth - 1)) - 1;
            else
                return countNodes(root->left) + countNodes(root->right) + 1;
        }
    }

    int getLeftDepth(TreeNode* root)
    {
        int count = 0;
        while (root != NULL)
        {
            count++;
            root = root->left;
        }
        return count;
    }

    int getRightDepth(TreeNode* root)
    {
        int count = 0;
        while (root != NULL)
        {
            count++;
            root = root->right;
        }
        return count;
    }
};