漫谈递归之完全二叉树
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题目 完全二叉树
分析
答案
//完全二叉树：
/**
1 Calculate the number of nodes (count) in the binary tree.
2 Start recursion of the binary tree from the root node of the binary tree with index (i) being set as 0 and the number of nodes in the binary (count).
3 If the current node under examination is NULL, then the tree is a complete binary tree. Return true.
4 If index (i) of the current node is greater than or equal to the number of nodes in the binary tree (count) i.e. (i>= count), then the tree is not a complete binary. Return false.
5 Recursively check the left and right sub-trees of the binary tree for same condition.
For the left sub-tree use the index as (2*i + 1) while for the right sub-tree use the index as (2*i + 2).
6 The time complexity of the above algorithm is O(n).
**/
bool isCompleteBinary(node *root)
{
int total=isCompleteBinary(root);
return is_complete_binary(root,1,total);
}
bool is_complete_binary(node *root,int index,int& length)
{
if(root ==NULL)
{
return true;
}
if(index>length)
{
return false;
}
return is_complete_binary(root->left,2*index,length) && is_complete_binary(root->right,2*index+1,length);
}
int get_node_number(node* root)
{
if(root ==NULL)
{
return 0;
}
return 1+get_node_number(root->left)+get_node_number(root->right);
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