[LeetCode] Count Complete Tree Nodes 求完全二叉树的节点个数

 

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.

 

这道题给定了一棵完全二叉树,让我们求其节点的个数。很多人分不清完全二叉树和满二叉树的区别,下面让我们来看看维基百科上对二者的定义:

完全二叉树:对于一颗二叉树,假设其深度为d(d>1)。除了第d层外,其它各层的节点数目均已达最大值,且第d层所有节点从左向右连续地紧密排列,这样的二叉树被称为完全二叉树;

满二叉树:对于上述的完全二叉树,如果去掉其第d层的所有节点,那么剩下的部分就构成一个满二叉树(此时该满二叉树的深度为d-1);

Image result for 满二叉树

 

通过上面的定义,我们可以看出二者的关系是,满二叉树一定是完全二叉树,而完全二叉树不一定是满二叉树。那么这道题给的完全二叉树就有可能是满二叉树,若是满二叉树,节点个数很好求,为2的h次方-1,h为该满二叉树的高度。这道题可以用递归和非递归两种方法来解。我们先来看递归的方法,思路是分别找出以当前节点为根节点的左子树和右子树的高度并对比,如果相等,则说明是满二叉树,直接返回节点个数,如果不相等,则节点个数为左子树的节点个数加上右子树的节点个数再加1(根节点),其中左右子树节点个数的计算可以使用递归来计算,参见代码如下:

 

class Solution {

public:

    int countNodes(TreeNode* root) {

        int hLeft = 0, hRight = 0;

        TreeNode *pLeft = root, *pRight = root;

        while (pLeft) {

            ++hLeft;

            pLeft = pLeft->left;

        }

        while (pRight) {

            ++hRight;

            pRight = pRight->right;

        }

        if (hLeft == hRight) return pow(2, hLeft) - 1;

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

    }

};

 

递归的解法还有一种解法如下所示:

 

class Solution {

public:

    int countNodes(TreeNode* root) {

        int hLeft = leftHeight(root);

        int hRight = rightHeight(root);

        if (hLeft == hRight) return pow(2, hLeft) - 1;

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

    }

    int leftHeight(TreeNode* root) {

        if (!root) return 0;

        return 1 + leftHeight(root->left);

    }

    int rightHeight(TreeNode* root) {

        if (!root) return 0;

        return 1 + rightHeight(root->right);

    }

};

 

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