LeetCode OJ：Lowest Common Ancestor of a Binary Search Tree（最浅的公共祖先）

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.

求最近的公共祖先，很显然，对于二叉搜索树来说，当两个节点的值都比root小的时候，lca应该在root左节点这一侧， 都比root值大的时候都在右节点这一侧，否则root

就是lca，代码见下所示：

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
         if(max(p->val, q->val) < root->val)
             lowestCommonAncestor(root->left, p, q);
         else if(min(p->val, q->val) > root->val)
             lowestCommonAncestor(root->right,p ,q);
         else
             return root;
     }
 };

而对于非二叉搜索树来说，做法一般是先分别查找到到P以及Q的路线，然后对比路线，找出LCA，代码如下：

 class Solution {
 public:
     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
         if(root == NULL || p == NULL || q == NULL)
             return NULL;
         vector<TreeNode *>pathP;
         vector<TreeNode *>pathQ;
         pathP.push_back(root);
         pathQ.push_back(root);
         getPath(root, p, pathP);
         getPath(root, q, pathQ);
         TreeNode * lca = NULL;
         for(int i = ; i < pathP.size() && i < pathQ.size(); ++i){
             if(pathP[i] == pathQ[i])　　//检查lca
                 lca = pathP[i];
             else break;
         }
         return lca;
     }

     bool getPath(TreeNode * root, TreeNode * target, vector<TreeNode * > & path)
     {
         if(root == target)
             return true;
         if(root->left){
             path.push_back(root->left);
             if(getPath(root->left, target, path)) return true;
             path.pop_back();
         }
         if(root->right){
             path.push_back(root->right);
             if(getPath(root->right, target, path)) return true;
             path.pop_back();
         }
         return false;
     }
 };

java版本的如下所示：

 public class Solution {
     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
         if(Math.max(p.val, q.val) < root.val){
             return lowestCommonAncestor(root.left, p, q);
         }else if(Math.min(p.val, q.val) > root.val){
             return lowestCommonAncestor(root.right, p, q);
         }else{
             return root;
         }
     }
 }