Leetcode中等：236. 二叉树的最近公共祖先

题目：二叉树的最近公共祖先

• 题号：236
• 难度：中等
• https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/

给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

例如，给定如下二叉树: root = [3,5,1,6,2,0,8,null,null,7,4]

示例 1:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
输出: 3
解释: 节点 5 和节点 1 的最近公共祖先是节点 3。

示例 2:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
输出: 5
解释: 节点 5 和节点 4 的最近公共祖先是节点 5。因为根据定义最近公共祖先节点可以为节点本身。

示例 3：

输入：root = [1,2], p = 1, q = 2
输出：1

提示:

• 树中节点数目在范围 $[2, 10^5]$ 内。
• $-10^9 <= Node.val <= 10^9$
• 所有 Node.val 互不相同 。
• p != q
• p 和 q 均存在于给定的二叉树中。

---

实现

思路：利用递归

C# 语言

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
 
public class Solution
{
    public TreeNode LowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q)
    {
        return Find(root, p, q);
    }

    private TreeNode Find(TreeNode current, TreeNode p, TreeNode q)
    {
        if (current == null || current == p || current == q)
            return current;
        TreeNode left = Find(current.left, p, q);
        TreeNode right = Find(current.right, p, q);
        if (left == null)
            return right;
        if (right == null)
            return left;
        return current;
    }
}

C++ 语言

/**
 * 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 == NULL || root == p|| root == q)
            return root;
        TreeNode* left =lowestCommonAncestor(root->left,p,q);
        TreeNode* right =lowestCommonAncestor(root->right,p,q);
        if(left == NULL)
            return right;
        if(right == NULL)
            return left;
        return root;        
    }
};

python 语言

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if root is None or root==p or root==q:
            return root
        l = self.lowestCommonAncestor(root.left,p,q)
        r = self.lowestCommonAncestor(root.right,p,q)
        if l is None:
            return r
        if r is None:
            return l
        return root