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.

最简单的方法s1：

/**

 * 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) {

        return count_nodes(root);

    }

    

    int count_nodes(TreeNode* root) {

        if (root == NULL) {

            return 0;

        }

        int cnt = 1;

        cnt += count_nodes(root->left) + count_nodes(root->right);

        return cnt;

    }

};

当然TLE了

改进后s2：

因为完全二叉树如果去掉最后最后一层，那么剩下的这H-1层组成的树就是一颗满二叉树，不用去数其中的节点，直接可以计算得出为(2^(H-1))-1个，所以只要求出最后一层的节点个数加上即可。

 

/**

 * 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) {

        if (root == NULL) {

            return 0;

        }

        int elen = 0;

        TreeNode* cur = root;

        

        while (cur != NULL) {

            elen++;

            cur = cur->left;

        }

        

        // amount of nodes above last level in the tree

        int cnt = (1<<(elen-1)) - 1;

        return cnt + last_level_count(root, elen);

        

    }

    

    int last_level_count(TreeNode* root, int elen) {

        if (root == NULL || elen <= 0) {

            return 0;

        }

        if (elen == 1) {

            return 1;

        }

        TreeNode* cur = root->left;

        int lcnt = elen - 1;

        while (cur != NULL) {

            lcnt--;

            cur = cur->right;

        }

        

        // left sub-tree is not full, no need to compute right sub-tree

        if (lcnt != 0) {

            return last_level_count(root->left, elen - 1);

        }

        // left sub-tree is full

        int level_count_left_subtree = 1<<(elen - 1 - 1);

        int level_count_right_subtree= last_level_count(root->right, elen - 1);

        

        return level_count_left_subtree + level_count_right_subtree;

    }

};

last_level_count函数可以考虑实现为非递归形式。

超简洁版本S3：

class Solution {



public:



    int countNodes(TreeNode* root) {



        if(!root) return 0;



        int hl=0, hr=0;



        TreeNode *l=root, *r=root;



        while(l) {hl++;l=l->left;}



        while(r) {hr++;r=r->right;}



        if(hl==hr) return pow(2,hl)-1;



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



    }



};

来自discuss，时间上和第二种一样