[LeetCode] 235. Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先

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:

2
8
6

Example 2:

2
4
2

Note:

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

求二叉搜索树(BST)的最近公共祖先(LCA)。最近公共祖先是指在一个树或者有向无环图中同时拥有v和w作为后代的最深的节点。

解法1：递归，

1. P, Q都比root小，则LCA在左树，我们继续在左树中寻找LCA

2. P, Q都比root大，则LCA在右树，我们继续在右树中寻找LCA

3. 其它情况，表示P,Q在root两边，或者二者其一是root，或者都是root，这些情况表示root就是LCA，直接返回root即可。

解法2: 迭代

判断标准同解法1，只是用迭代来实现。

Java:

public 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);
        if(root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q);
        return root;
    }
}

Java:

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        //发现目标节点则通过返回值标记该子树发现了某个目标结点
        if(root == null || root == p || root == q) return root;
        //查看左子树中是否有目标结点，没有为null
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        //查看右子树是否有目标节点，没有为null
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        //都不为空，说明做右子树都有目标结点，则公共祖先就是本身
        if(left!=null&&right!=null) return root;
        //如果发现了目标节点，则继续向上标记为该目标节点
        return left == null ? right : left;
    }
}　　

Python:

class Solution:
    # @param {TreeNode} root
    # @param {TreeNode} p
    # @param {TreeNode} q
    # @return {TreeNode}
    def lowestCommonAncestor(self, root, p, q):
        s, b = sorted([p.val, q.val])
        while not s <= root.val <= b:
            # Keep searching since root is outside of [s, b].
            root = root.left if s <= root.val else root.right
        # s <= root.val <= b.
        return root

C++:

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

C++:

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