235. Lowest Common Ancestor of a Binary Search Tree(unsolved)

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.

思路：
注意到二叉搜索树的性质，即左子树的值小于根节点的值，根节点小于右节点。那么假如两个节点都小于根节点，就进入左子树考察，都大于就右子树。假如是一大一小，那么就返回根节点，因为这时分开了就说明是一个在左边一个在右边了。应注意到这里的返回值是递归函数，那么就可以逐层直接返回了，要学习这一点。

/** * 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(p->val<root->val&&q->val<root->val)            return lowestCommonAncestor(root->left,p,q);        if(p->val>root->val&&q->val>root->val)            return lowestCommonAncestor(root->right,p,q);        return root;    }};