题目描述：
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.
利用二叉排序树的性质，可以很好地解决这道题。
solution：
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if (root == NULL || p == NULL || q == NULL) return NULL; if (root->val > p->val && root->val > q->val) return lowestCommonAncestor(root->left, p, q); else if (root->val < p->val && root->val < q->val) return lowestCommonAncestor(root->right, p, q); else return root; }
如果只是一颗普通的二叉树呢？
solution：
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if (!root) return NULL; if (root == p || root == q) return root; TreeNode *L = lowestCommonAncestor(root->left, p, q); TreeNode *R = lowestCommonAncestor(root->right, p, q); if (L && R) return root; return L ? L : R; }
来源：Lowest Common Ancestor of a Binary Tree Part I
如果二叉树的结点存在指向父结点的指针，问题可以转化为求两个单链表的相交结点。
solution：
struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode *parent; }; int getHeight(TreeNode *p) { int height = 0; while (p) { height++; p = p->parent; } return height; } TreeNode* lowestCommonAncestor(TreeNode *p, TreeNode *q) { int h1 = getHeight(p); int h2 = getHeight(q); if (h1 > h2) { swap(h1, h2); swap(p, q); } int dh = h2 - h1; for (int h = 0; h < dh; ++h) q = q->parent; while (p && q) { if (p == q) return p; p = p->parent; q = q->parent; } return NULL; }