[LeetCode]235. Lowest Common Ancestor of a Binary Search Tree

235. 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.

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递归版本

/**
 * 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) {
        // 前提：BST中没有相同的值


        if(root == NULL || p == NULL || q == NULL)
            return NULL;

        // p和q都在左子树
        if(p->val < root->val && q->val < root->val)
            return lowestCommonAncestor(root->left, p, q);

        // p和q都在右子树
        if(p->val >root->val && q->val >root->val)
            return lowestCommonAncestor(root->right, p, q);

        // q和q如果不在同一子树，哈哈，那root就是它们的最低公共祖先了
        return root;
    }
};

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迭代版本

/**
 * 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) {
        TreeNode* pCur = root;
        while(true){
            if(p->val < pCur->val && q->val < pCur->val)
                pCur = pCur->left;
            else if(p->val > pCur->val && q->val > pCur->val)
                pCur = pCur->right;
            else
                return pCur;
        }
        // return pCur;
    }
};