leetcode解题: Lowest Common Ancestor of a Binary Search Tree (235)

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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.

解法1：Recursion

这题有一个很重要的前提是这是一棵BST。并且隐含条件是一定存在对于给点的两个node的LCA。
那么考虑两种情况，假设两个node的大小一个比root大一个比root小，则node分属于左右子树，LCA只可能是root
如果两个node同属一边，则问题也等价于在左子树（假设node的val比root的小），套用递归的思想解决即可。
C++

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/**
 * 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 (root == p || root == q) {
            return root;
        }
        if (p->val < root->val && q->val > root->val) {
            return root;
        }
        if (p->val > root->val && q->val < root->val) {
            return root;
        }
        if (p->val < root->val) {
            return lowestCommonAncestor(root->left, p, q);
        }
        return lowestCommonAncestor(root->right, p, q);
    }
};

Java

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        
        if (root == p || root == q) {
            return root;
        }
        
        if ((root.val < p.val && root.val > q.val) || (root.val > p.val && root.val < q.val)) {
            return root;
        }
        
        if (root.val < p.val && root.val < q.val) {
            return lowestCommonAncestor(root.right, p, q);
        }
        
        return lowestCommonAncestor(root.left, p, q);
        
    }
}