# Concise Elegant C++ Solution Beats 99% With Detailed Comments

• ``````/* since the tree is complete the height of the tree */
/* is the height of the left most node */
int HeightCompleteTree(TreeNode * root) {
if (!root) { return 0; }
return HeightCompleteTree(root->left) + 1;
}

int countNodes(TreeNode* root) {
if (!root) { return 0; }
int cur_count = 1;
TreeNode* next = root;
/* just like binary search, during each iteration we get a reference of the */
/* root of the subtree where the LAST NODE of the complete binary tree resides in */
/* while we are doing that we also calculate the current node count as if that node is */
/* the last node, the iteration terminates when the root of the subtree is the last node itself,
* by that time we know the count of the nodes already */
while (next->left || next->right) {

/* this means the LAST NODE is at the left subtree */
if (HeightCompleteTree(next->left) > HeightCompleteTree(next->right)) {
next = next->left;
/* update the current count */
cur_count = cur_count << 1;
} else {
next = next->right;
/* update the current count */
cur_count = (cur_count << 1) + 1;
}

}

/* while loop terminate when "next" is the LAST NODE */
return cur_count;
}
``````

• Elegant codes, thanks for sharing~

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