[leetcode] 222. Count Complete Tree Nodes 解题报告

题目链接：https://leetcode.com/problems/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.

思路：求完全二叉树的结点个数。可以求左右子树的高度，

１．如果相等则说明左子树是满二叉树，那么可以根据左子树的高度求出左子树加上根节点的结点数为2^leftHeight，再加上右子树的结点即为以当前结点为根的结点总数

２．否则说明右子树为满二叉树，同样右子树的结点加上根节点的结点数为2^rightHeight，再加上左子树结点即为当前结点为根的结点总数

代码如下：

/**
 * 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 getHeight(TreeNode* root)
    {
        if(!root) return 0;
        return 1 + getHeight(root->left);
    }
    int countNodes(TreeNode* root) {
        if(!root) return 0;
        int lh = getHeight(root->left);
        int rh = getHeight(root->right);
        if(lh == rh)
            return pow(2, lh) + countNodes(root->right);
        else
            return pow(2, rh) + countNodes(root->left);
    }
};

参考：https://leetcode.com/discuss/73892/a-very-clear-recursive-solution-isnt-it