//二叉查找樹的第k大節(jié)點(diǎn)：利用二叉搜索樹的中序遍歷序列是遞增排序的性質(zhì)，遞歸實(shí)現(xiàn) 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); } //二叉樹的深度：二叉樹的后序遍歷 //題目一：求二叉樹的深度 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; }//題目二：判斷一棵樹是否是二叉平衡樹 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); } //后序遍歷的方法：每個節(jié)點(diǎn)只需要遍歷一次 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); }

轉(zhuǎn)載于:https://www.cnblogs.com/songdanzju/p/7441971.html

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