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;        }    }};