【LeetCode】235. Lowest Common Ancestor of a Binary Search Tree

问题描述

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.

寻找两个节点的公共最低祖先节点。

问题分析

要注意这不止是一棵二叉树,更是一棵二叉搜索树。也就是说,里面每个节点的值都不一样,并且是有顺序的。所以,可以有如下算法:

根节点root,寻找节点A和B的最低公共祖先节点:
首先判断root的值是不是就是A或者B,如果是,直接返回A或者B;
然后判断A和B是不是一个在root的左边,一个在root的右边,如果是,那就返回root;
再判断A和B是不是都在root的左边,如果是,就令root=root->left,继续下轮寻找;
最后判断A和B是不是都在root的右边,如果是,就令root=root->right,继续下轮寻找。

代码

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root==0) return 0;
        if(p==0||q==0) return root;
        int val = root->val;
        if(p->val==val) return p;
        if(q->val==val) return q;
        if(p->val<val&&q->val>val||p->val>val&&q->val<val) {
            return root;
        }
        if(p->val<val&&q->val<val) return lowestCommonAncestor(root->left, p, q);
        if(p->val>val&&q->val>val) return lowestCommonAncestor(root->right, p, q);
    }
};