如何递归检查给定的红黑树是否遵守&#34规则;从节点到空链路的每条路径必须包含相同数量的黑色节点"。 我使用这种结构:
enum color = {RED, BLACK};
typedef struct node {
int value;
struct node* left;
struct node* right;
color c;
} node;
我尝试使用此原型实现算法:
bool isRBT(struct node* tree, int numberBlackNodesLeft, int numberBlackNodesRight)
但是,我无法递归计算这些数字。因为,规则强制要求来自一个节点的每条路径都必须重复该规则。这对我来说很难实现。
有什么好主意吗?
提前致谢!
答案 0 :(得分:6)
这是一种简单的方法:
// Returns the number of black nodes in a subtree of the given node
// or -1 if it is not a red black tree.
int computeBlackHeight(node* currNode) {
// For an empty subtree the answer is obvious
if (currNode == NULL)
return 0;
// Computes the height for the left and right child recursively
int leftHeight = computeBlackHeight(currNode->left);
int rightHeight = computeBlackHeight(currNode->right);
int add = currNode->color == BLACK ? 1 : 0;
// The current subtree is not a red black tree if and only if
// one or more of current node's children is a root of an invalid tree
// or they contain different number of black nodes on a path to a null node.
if (leftHeight == -1 || rightHeight == -1 || leftHeight != rightHeight)
return -1;
else
return leftHeight + add;
}
要测试树是否满足红黑树的黑色高度属性,您可以将上述函数包装为:
bool isRBTreeBlackHeightValid(node* root)
{
return computeBlackHeight(root) != -1;
}