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.
分析：
在二叉查找树种，寻找两个节点的最低公共祖先。
1、如果a、b都比根节点小，则在左子树中递归查找公共节点。此时对root->left递归求解
2、如果a、b都比根节点大，则在右子树中查找公共祖先节点。对root->right递归求解
3、如果a、b一个比根节点大，一个比根节点小，或者有一个等于根节点，则根节点即为最低公共祖先。
c递归代码：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * struct TreeNode *left; 6 * struct TreeNode *right; 7 * }; 8 */ 9 struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) { 10 if(root == NULL || p == NULL || q == NULL) 11 return NULL; 12 if(root->val > p->val && root->val > q->val){ 13 return lowestCommonAncestor(root->left, p, q); 14 } 15 else if(root->val < p->val && root->val < q->val){ 16 return lowestCommonAncestor(root->right, p, q); 17 } 18 else{ 19 return root; 20 } 21 }
c非递归代码：
1 struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) { 2 if(root == NULL || p == NULL || q == NULL) 3 return NULL; 4 while(root != NULL){ 5 if(root->val > p->val && root->val > q->val) 6 root = root->left; 7 else if(root->val < p->val && root->val < q->val) 8 root = root->right; 9 else 10 break; 11 } 12 return root; 13 }
C++代码：
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 == NULL || p == NULL || q == NULL) 14 return NULL; 15 int min1 = min(p->val, q->val); 16 int max1 = max(p->val, q->val); 17 //if((root->val >= p->val && root->val <= q->val) || (root->val >= q->val && root->val <= p->val)) 18 // return root; 19 if(max1 < root->val){ 20 return lowestCommonAncestor(root->left, p, q); 21 } 22 else if(min1 > root->val){ 23 return lowestCommonAncestor(root->right, p, q); 24 } 25 else{ 26 return root; 27 } 28 } 29 };
leetcode编译器运行，第一个程序用时最少，依次递增。
除了这种解法，也可以把它当做普通的二叉树做。