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：

 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(max(p->val, q->val) < root->val)
14             lowestCommonAncestor(root->left, p, q);
15         else if(min(p->val, q->val) > root->val)
16             lowestCommonAncestor(root->right,p ,q);
17         else 
18             return root;       
19     }
20 };

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：

 1 class Solution {
 2 public:
 3     TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
 4         if(root == NULL || p == NULL || q == NULL) 
 5             return NULL;
 6         vector<TreeNode *>pathP;
 7         vector<TreeNode *>pathQ;
 8         pathP.push_back(root);
 9         pathQ.push_back(root);
10         getPath(root, p, pathP);
11         getPath(root, q, pathQ);
12         TreeNode * lca = NULL;
13         for(int i = 0; i < pathP.size() && i < pathQ.size(); ++i){
14             if(pathP[i] == pathQ[i])　　//检查lca
15                 lca = pathP[i];
16             else break;
17         }
18         return lca;
19     }
20 
21     bool getPath(TreeNode * root, TreeNode * target, vector<TreeNode * > & path)
22     {
23         if(root == target)
24             return true;
25         if(root->left){
26             path.push_back(root->left);
27             if(getPath(root->left, target, path)) return true;
28             path.pop_back();
29         }
30         if(root->right){
31             path.push_back(root->right);
32             if(getPath(root->right, target, path)) return true;
33             path.pop_back();
34         }
35         return false;
36     }
37 };

java版本的如下所示：

 1 public class Solution {
 2     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
 3         if(Math.max(p.val, q.val) < root.val){
 4             return lowestCommonAncestor(root.left, p, q);
 5         }else if(Math.min(p.val, q.val) > root.val){
 6             return lowestCommonAncestor(root.right, p, q);
 7         }else{
 8             return root;
 9         }
10     }
11 }

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原文地址：https://www.cnblogs.com/-wang-cheng/p/4905291.html