235. Lowest Common Ancestor of a Binary Search Tree

问题描述

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.

题目链接：

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思路分析

给一个二叉查找树和它的两个结点，找到两个结点的最低共同祖先。根据二叉搜索树的性质，左子树元素小于root小于右子树元素，而且左右子树都是二叉搜索树。可以递归查找，找到将两个结点分到左右子树的那时的root结点即可。

代码

递归

/**
 * 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;
    }
};

时间复杂度：$O(logn)$
空间复杂度：$O(1)$

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反思

也可以不递归，用一个指针来便利即可。

class Solution { 
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        TreeNode* cur = root;
        while (true) {
            if (p -> val < cur -> val && q -> val < cur -> val)
                cur = cur -> left;
            else if (p -> val > cur -> val && q -> val > cur -> val)
                cur = cur -> right;
            else return cur; 
        }
    }
};