leetcode刷题，总结，记录，备忘235

leetcode235

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.
不算难的题目，之前像太复杂了，，，没绕过弯，看了别人的代码，突然就开朗了，之前自己还是有点蠢。根据根节点的值域和p，q两个节点的值域进行判断，如果都比根的节点值域小，就代表在左子树中，用根节点的左节点做递归，反之亦然。当2个节点的值域一个大于根节点的值域，一个小于的时候，代表找到，即返回。。。这么简单，我之前竟然蠢了好久，，，诶，，

/**
 * 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 == NULL || p == NULL || q == NULL) {
            return NULL;
        }
        
        if (max(p->val, q->val) < root->val) {
            return lowestCommonAncestor(root->left, p, q);
        } else if (min(p->val, q->val) > root->val) {
            return lowestCommonAncestor(root->right, p, q);
        } else {
            return root;
        }
    }
};