[Leetcode] 235. Lowest Common Ancestor of a Binary Search Tree 解题报告

题目：

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：

由于是二叉搜索树，所以左子树的所有结点的值都要比根结点小，右子树的所有结点的值都要比根结点大。利用这一特性，我们可以递归实现求解最低公共祖先的算法：假设p的结点值小于q，那么如果p和q分别位于root的左右子树上，则root必然就是p和q的最低公共祖先；否则如果p和q的都位于root的右子树上，那么p和q的最低公共祖先必然位于root的右子树上；否则p和q的最低公共祖先必然位于root的左子树上。后两种情况可以通过递归求解。

代码：

/**
 * 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) {
        TreeNode *smaller = p->val < q->val ? p : q;
        TreeNode *larger = p->val < q->val ? q : p;
        if (root->val >= smaller->val && root->val <= larger->val) {
            return root;
        }
        else if (root->val < smaller->val) {
            return lowestCommonAncestor(root->right, smaller, larger);
        }
        else {
            return lowestCommonAncestor(root->left, smaller, larger);
        }
    }
};