二叉搜索树相关题目

//二叉查找树的第k大节点：利用二叉搜索树的中序遍历序列是递增排序的性质，递归实现
struct BinaryTreeNode{int m_nValue;BinaryTreeNode* m_pLeft;BinaryTreeNode* m_pRight;
};
BinaryTreeNode* FindKthNodeCore(BinaryTreeNode* pRoot, unsigned int & k){BinaryTreeNode* target = nullptr;if (pRoot->m_pLeft != nullptr) {target = FindKthNodeCore(pRoot->m_pLeft, k);}if (target == nullptr){if (k == 1)target = pRoot;else--k;}if (target == nullptr && pRoot->m_pRight != nullptr){target = FindKthNodeCore(pRoot->m_pRight, k);}return target;
}
BinaryTreeNode* FindKthNode(BinaryTreeNode* pRoot, unsigned int k){if (pRoot == nullptr || k < 1)return nullptr;return FindKthNodeCore(pRoot, k);
}

//二叉树的深度：二叉树的后序遍历
//题目一：求二叉树的深度
int DepthOfTree(BinaryTreeNode* pRoot){if (pRoot == nullptr)return 0;int left = DepthOfTree(pRoot->m_pLeft);int right = DepthOfTree(pRoot->m_pRight);int depth = max(left, right) + 1;return depth;
}//题目二：判断一棵树是否是二叉平衡树
int DepthOfNode(BinaryTreeNode* pRoot){if (pRoot == nullptr)return 0;int left = DepthOfNode(pRoot->m_pLeft);int right = DepthOfNode(pRoot->m_pRight);return max(left, right) + 1;
}
bool IsBalancedTree(BinaryTreeNode* pRoot){if (pRoot == nullptr)return false;int left = DepthOfNode(pRoot->m_pLeft);int right = DepthOfNode(pRoot->m_pRight);int depthDiff = left - right;if (right > left)depthDiff = right - left;if (depthDiff > 1)return false;return IsBalancedTree(pRoot->m_pLeft) && IsBalancedTree(pRoot->m_pRight);
}
//后序遍历的方法：每个节点只需要遍历一次
bool IsBalanced(BinaryTreeNode* pRoot, int& depth){if (pRoot == nullptr){depth = 0;return true;}int left, right;if (IsBalanced(pRoot->m_pLeft, left)&& IsBalanced(pRoot->m_pRight, right)){int diff = left - right;if (diff >= -1 && diff <= 1){depth = max(left, right) + 1;return true;}}return false;
}
bool IsBalancedTree2(BinaryTreeNode* pRoot){int depth = 0;return IsBalanced(pRoot, depth);
}