235. Lowest Common Ancestor of a Binary Search Tree 树面试试题---最低公共祖先

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#程序员面试攻略 #二元搜索树 #最低公共祖先

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本文探讨了在二叉搜索树中查找两个指定节点的最低公共祖先（LCA）的问题，提供了两种解决方案：一种是非递归方法，通过循环进行节点比较；另一种是递归方法，效率更高。示例展示了不同输入情况下的正确输出。

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself 
             according to the LCA definition.

Note:

All of the nodes' values will be unique.
p and q are different and both values will exist in the BST.

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) {
        /*//非递归，因为有while循环，所以用的时间比较多
        TreeNode *curNode = root;
        while(1){
            if(curNode->val > p->val && curNode->val > q->val)
                curNode = curNode->left;
            else if(curNode->val < p->val && curNode->val < q->val)
                curNode = curNode->right;
            else
                return curNode;
        }*/
        //递归，效率高
        if (root->val > p->val && root->val > q->val) 
            return lowestCommonAncestor(root->left, p, q);
        else if (root->val < p->val && root->val < q->val) 
            return lowestCommonAncestor(root->right, p, q);
        else return root;
    }
};