235. Lowest Common Ancestor of a Binary Search Tree

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• Difficulty: Easy
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Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the : “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).”

_______6______       /              \    ___2__          ___8__   /      \        /      \   0      _4       7       9         /  \         3   5

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.

Solution 1: dfs 这是lowest ancestor of binary tree的做法，BST: left->val < root->val < right->val 可以利用这个

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 || p == root || q == root) return root;14         TreeNode* left = lowestCommonAncestor(root -> left, p, q);15         TreeNode* right = lowestCommonAncestor(root -> right, p, q);16         if(left && right) return root;17         return left ? left : right;18     }19 };

Solution 2: 递归

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 && p -> val < root -> val && q -> val > root -> val) return root;14         //TreeNode* left = lowestCommonAncestor(root -> left, p, q);15         //TreeNode* right = lowestCommonAncestor(root -> right, p, q);16         if(!root) return NULL;17         if(root -> val > max(p -> val, q -> val)){18             return lowestCommonAncestor(root -> left, p, q);//如果root值同时大于他俩，则要往左找19         }20         else if(root -> val < min(p -> val, q -> val)){21             return lowestCommonAncestor(root -> right, p, q);//如果root值同时xiao于他俩(min)，则要往right找22         }23         else return root;24     }25 };

 

Solution 3：while

同理，只不过root一直在走（更新）

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         while(true){15             if(root -> val > max(p -> val, q -> val)){16                 root = root -> left;17             }else if(root -> val < min(p -> val, q -> val)){18                 root = root -> right;19             }else{20                 break;21             }22         }23         return root;24     }25 };