Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
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For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
Analyse: For a binary search tree, we can recursively compare the node value and p, q value.
Trial 1: comparison
Runtime: 44ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { 13 if(!root) return NULL; 14 if(root == p || root == q) return root; 15 if(root->val < p->val && root->val < q->val) return lowestCommonAncestor(root->right, p, q); 16 if(root->val > p->val && root->val > q->val) return lowestCommonAncestor(root->left, p, q); 17 return root; 18 } 19 };
Trial 2: same solution as Lowest Common Ancestor.
Runtime: 44ms.
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { 13 if(!root) return NULL; 14 if(root == p || root == q) return root; 15 TreeNode* leftSub = lowestCommonAncestor(root->left, p, q); 16 TreeNode* rightSub = lowestCommonAncestor(root->right, p, q); 17 if(leftSub && rightSub) return root; 18 return leftSub ? leftSub : rightSub; 19 } 20 };