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 = 0; 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;
        }
    }
}