[LeetCode] 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.

解题思路

全然遍历会超时。能够从根节点開始分别计算左子树和右子树的高度。假设相等，则为满二叉树，结点个数为$2^n-1$。假设高度不相等，则结点个数为左子树结点个数+右子树结点个数+1。（高度n是包括根节点的高度）

实现代码

// Runtime: 100 ms
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
class Solution {
public:
    int countNodes(TreeNode* root) {
        int leftHight = getLeft(root);
        int rightHeight = getRight(root);
        if (leftHight == rightHeight)
        {
            return pow(2, leftHight) - 1;
        }

        return 1 + countNodes(root->left) + countNodes(root->right);
    }

private:
    int getLeft(TreeNode *root)
    {
        int height = 0;
        while (root)
        {
            height++;
            root = root->left;
        }

        return height;
    }

    int getRight(TreeNode *root)
    {
        int height = 0;
        while (root)
        {
            height++;
            root = root->right;
        }

        return height;
    }
};