#235 Lowest Common Ancestor of a Binary Search Tree
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).”
_______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.
The best code so far:
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 TreeNode *cur = root; 14 while (cur != NULL) { 15 if (cur->val > p->val && cur->val > q->val) cur = cur->left; 16 else if (cur->val < p->val && cur->val < q->val) cur = cur->right; 17 else return cur; 18 } 19 return cur; 20 } 21 };
http://www.cnblogs.com/easonliu/p/4639043.html
不用迭代效率显然要高啊！
Code by S:
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 /*TreeNode* tmp; 16 if (p->val == q->val) 17 return p; 18 if (p->val > q->val){ 19 tmp = p; 20 p = q; 21 q = tmp; 22 delete tmp; 23 }*/ 24 25 if((p->val < root->val) && (q->val < root->val)){ 26 root = root->left; 27 return lowestCommonAncestor(root, p, q); 28 } 29 else if((p->val > root->val) && (q->val > root->val)){ 30 root = root->right; 31 return lowestCommonAncestor(root, p, q); 32 } 33 else 34 return root; 35 } 36 };
默认p->val <= q->val; 一开始没考虑NULL的情况，以及一开始迭代初每次判断一左一右则终止这个，一直报runtime error的错，最后把这个判断放在else里面就跑过了。
#237 Lowest Common Ancestor of a Binary Tree
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
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).”
_______3______
/ ___5__ ___1__
/ \ / 6 _2 0 8
/ 7 4
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
The best code so far
http://wp.javayu.me/2014/02/lowest-common-ancestor-of-a-binary-tree/
http://articles.leetcode.com/2011/07/lowest-common-ancestor-of-a-binary-tree-part-i.html
http://blog.csdn.net/xudli/article/details/8560362
Code by S:
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 if(root == p || root == q) 16 return root; 17 TreeNode *L = lowestCommonAncestor(root->left, p, q); 18 TreeNode *R = lowestCommonAncestor(root->right, p, q); 19 if(L && R) 20 return root; 21 return L ? L : R; 22 } 23 };
想通道理都花了好久- -，被自己蠢到了。
原文：http://www.cnblogs.com/leetcodegroup/p/4654116.html