LeetCode 235. Lowest Common Ancestor of a Binary Search Tree

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本文介绍如何在二叉搜索树中找到两个指定节点的最近公共祖先，并详细解释了二叉搜索树的基本特性及算法实现。

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 allowa 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.

It is important to understand the basic feature of BST.

   TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root->val == p->val || root->val == q->val) return root;
        else if(p->val > root->val && q->val > root->val) return lowestCommonAncestor(root->right, p, q);
        else if(p->val < root->val && q->val < root->val) return lowestCommonAncestor(root->left, p, q);
        else return root;
    }

Find the lowest common ancestor of the given two targets, consider the case that the two targets are not in the BST.

// lowest common ancestor of binary search tree.
bool find(TreeNode* root, int target) {
  if(!root) return false;
  if(root->val == target) return true;
  else if(root->val < target) find(root->left, target);
  else find(root->right, target);
}

TreeNode* lowestCommonAncestor(TreeNode* root, int v1, int v2) {
  if(!root || root->val == v1 && root->val == v2) return NULL;
  if(root->data > x && root->val > y) {
    return lowestCommonAncestor(root->left, v1, v2);
  } else if(root->val < x && root->val < y) {
    return lowestCommonAncestor(root->right, v1, v2);
  } else if(find(root, v1) && find(root, v2)) return root;
  else return NULL;
}