leetcode 236.二叉树的最近公共祖先_cwtnice的博客-爱代码爱编程

原题链接：Leetcode 236. Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

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).”

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 10⁵].
• -10⁹ <= Node.val <= 10⁹
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

方法一：递归

思路：

C++代码：

/**
 * 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:

    /*
    函数的功能:
    给定两个节点 p 和 q
    1. 如果 p 和 q 都存在，则返回它们的公共祖先
    2. 如果只存在一个，则返回存在的一个
    3. 如果 p 和 q 都不存在，则返回NULL
    */
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        // 递归的出口
        if(root == NULL)
            return NULL;

        // 如果只存在一个，则返回存在的一个
        if(root == p || root == q) 
            return root;
            
        // 可认为已经实现了左右子树算出的结果
        TreeNode* left =  lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
       
        // 如果有一个为空 那么答案只看另一个
        if(left == NULL)
            return right;
        if(right == NULL)
            return left;

        // p和q在两侧 此时的root就是结果
        if(left && right) 
            return root;
        
        // 必须有返回值
        return NULL; 
    }
};