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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 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.
思路：
利用BST的性质，若待查找两个结点的数值都小于当前结点则向左边查找（若有一个等于当前结点则当前结点就是其最小的公共结点，结束查找），反之向右边查找（同理），若一个大于一个小于，则当前结点就是其最小的公共结点，结束查找。
C++：
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 private: 12 TreeNode *l, *r, *ret; 13 public: 14 15 void rec(TreeNode *root) 16 { 17 int v = root->val; 18 if(l->val <= v && r->val <= v) 19 { 20 if(v == l->val || v == r->val) 21 { 22 ret = root; 23 return ; 24 } 25 26 if(root->left != 0) 27 rec(root->left); 28 } 29 else if(l->val < v && v < r->val) 30 { 31 ret = root; 32 return ; 33 } 34 else if(l->val >= v && r->val >= v) 35 { 36 if(v == l->val || v == r->val) 37 { 38 ret = root; 39 return ; 40 } 41 42 if(root->right != 0) 43 rec(root->right); 44 } 45 } 46 47 TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { 48 if(root == 0) 49 return 0; 50 51 if(p->val < q->val){ 52 l = p; 53 r = q; 54 } 55 else{ 56 l = q; 57 r = p; 58 } 59 60 ret = 0; 61 62 rec(root); 63 64 return ret; 65 } 66 };
【LeetCode 235】Lowest Common Ancestor of a Binary Search Tree
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原文地址：http://www.cnblogs.com/tjuloading/p/4641093.html
