二叉树中任意两个节点的最近公共祖先

public class Solution {  
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {  
        //发现目标节点则通过返回值标记该子树发现了某个目标结点  
        if(root == null || root == p || root == q) return root;  
        //查看左子树中是否有目标结点，没有为null  
        TreeNode left = lowestCommonAncestor(root.left, p, q);  
        //查看右子树是否有目标节点，没有为null  
        TreeNode right = lowestCommonAncestor(root.right, p, q);  
        //都不为空，说明做右子树都有目标结点，则公共祖先就是本身  
        if(left!=null&&right!=null) return root;  
        //如果发现了目标节点，则继续向上标记为该目标节点  
        return left == null ? right : left;  
    }  
}  

思路：从根节点开始遍历，如果node1和node2中的任一个和root匹配，那么root就是最低公共祖先。 如果都不匹配，则分别递归左、右子树，如果有一个 节点出现在左子树，并且另一个节点出现在右子树，则root就是最低公共祖先.  如果两个节点都出现在左子树，则说明最低公共祖先在左子树中，否则在右子树。

感觉很奇妙。引申的问题

如果给定的不是二叉树，而是二叉搜索树呢？会比较简单一点，如果是带有指向父节点的指针的树，可以转化为两个链表求交汇点的问题。

 235. Lowest Common Ancestor of a Binary Search Tree

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.

/**
 * 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) {
        while((root -> val - p -> val)*(root -> val - q -> val) > 0 ){
            root = root -> val > p -> val ? root -> left : root -> right;
        }
        return root;
    }
};