给定一棵红黑树,我需要编写一个有效的算法来检查对于每个节点,从节点到后代叶子的所有路径是否包含相同数量的黑色节点,即如果属性为 true 或 false,则算法应返回布尔值。
它将返回RB树的黑色高度。如果高度为 0,则该树是无效的红色黑色树。
int BlackHeight(NodePtr root)
{
if (root == NULL)
return 1;
int leftBlackHeight = BlackHeight(root->left);
if (leftBlackHeight == 0)
return leftBlackHeight;
int rightBlackHeight = BlackHeight(root->right);
if (rightBlackHeight == 0)
return rightBlackHeight;
if (leftBlackHeight != rightBlackHeight)
return 0;
else
return leftBlackHeight + root->IsBlack() ? 1 : 0;
}
下面的代码验证沿着任何路径的黑色节点的数量是否相同
#define RED 0
#define BLACK 1
struct rbt {
int data;
struct rbt *left;
struct rbt *right;
int parent_color; // parent link color
uint64_t nodes_count; // to store number of nodes present including self
int level; // to store level of each node
};
typedef struct rbt rbt_t;
int check_rbt_black_height(rbt_t *root)
{
if(!root) return 1;
else {
int left_height = check_black_violation(root->left);
int right_height = check_black_violation(root->right);
if (left_height && right_height && left_height != right_height) {
printf("Error: Black nodes are not same @level=%d\n", root->level);
exit(1);
}
if (left_height && right_height)
// do not increment count for red
return is_color_red(root) ? left_height : left_height + 1;
else
return 0;
}
}
int is_color_red(rbt_t *root)
{
if(root && root->parent_color == RED) return 1;
return 0;
}