235. Lowest Common Ancestor of a Binary Search Tree

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• Total Accepted: 129842
• Total Submissions: 337481
• Difficulty: Easy
• Contributor: LeetCode

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.

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这一题的意思是要我们寻找BST两个节点p、q的最小共同根

我想到的第一个方法是，先保存从根到p节点上的所有点，包括p自己，保存在一个数组v中。然后从根节点开始寻找q，在寻找q的过程中，经过的最后一个包含在v中的节点就是他们的共同节点。这样的方法时间复杂度为O(logm*n)，空间复杂度为O(logm)。于是我开始寻找更好的方法，尽量不需要额外空间。经过思考后，想出了一个主意。可以先从根节点开始同时寻找p和q，一定有一段路径是它们共同的。只要沿着这条路走，当它们到达某个节点时总会“分叉”开。这个"分叉路口”节点就是它们共同最小根节点。

代码如下：

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)        //共同路径
                cur = cur -> left;
            else if (p -> val > cur -> val && q -> val > cur -> val)  //共同路径
                cur = cur -> right;
            else return cur;    //分叉点
        }
    }
};