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

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 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]

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.

解析

这道题我们可以用递归来求解，我们首先来看题目中给的例子，由于二叉搜索树的特点是左<根<右，所以根节点的值一直都是中间值，大于左子树的所有节点值，小于右子树的所有节点值，那么我们可以做如下的判断，如果根节点的值大于p和q之间的较大值，说明p和q都在左子树中，那么此时我们就进入根节点的左子节点继续递归，如果根节点小于p和q之间的较小值，说明p和q都在右子树中，那么此时我们就进入根节点的右子节点继续递归，如果都不是，则说明当前根节点就是最小共同父节点，直接返回即可。

解法1：递归

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        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;
    }
};

解法2：非递归

当然，此题也有非递归的写法，用个while循环来代替递归调用即可，然后不停的更新当前的根节点，也能实现同样的效果。

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        while(1){
            if (root->val > max(p->val, q->val)) root = root->left;
            else if (root->val < min(p->val, q->val)) root = root->right;
            else break;
        }
        return root;
    }
};

参考

http://www.cnblogs.com/grandyang/p/4640572.html