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