LeetCode 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.

给定一课平衡二叉树（BST），找到给定的两个节点的最近共同祖先（LCA）

根据definition of LCA on Wikipedia（LCA在Wiki上的定义），最近共同祖先被定义为后代包括v和w的最近的节点T（一个节点的后代也可以包含它本身）

        _______6______       /              \    ___2__          ___8__   /      \        /      \   0      _4       7       9         /  \         3   5

例如，在这棵平衡二叉树中，节点2和8的最近共同祖先为节点6，而节点2和4的最近共同祖先为节点2，因为一个节点的后代也包含它自己。

百度了别人的之后仿照写的。。。自己写的虽然思路相同，可是死活运行错误。。。心好累←_←并不清楚原因

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