题目要求

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 的两个节点 p、q，最近公共祖先表示为一个节点 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）。”

ACM

本题适合从下往上遍历，所以使用后序遍历来递归。

#include <iostream>
#include <vector>
using namespace std;
// #include <unordered_map>
// #include <algorithm>

struct TreeNode
{
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode* left, TreeNode* right) : val(x), left(left), right(right) {}
};

class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
    {
        if (root == p || root == q || root == NULL)
        {
            return root;
        }    

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

int main(void)
{
    TreeNode* root = new TreeNode(8);
    root->left = new TreeNode(10);
    root->right = new TreeNode(4);
    root->left->left = new TreeNode(1);
    root->left->right = new TreeNode(7);
    root->right->left = new TreeNode(15);
    root->right->right = new TreeNode(20);
    root->left->right->left = new TreeNode(6);
    root->left->right->right = new TreeNode(5);
    Solution solution;

    TreeNode* p = root->left->right->left;
    TreeNode* q = root->left->right->right;
    TreeNode* result = solution.lowestCommonAncestor(root, p, q);
    cout << "Lowest Common Ancestor: " << result->val << endl;
    return 0;   
}

测试代码中 p、q 的定义，不能简单地定义一个根节点，TreeNode* p = new TreeNode(6);

TreeNode* p = new TreeNode(5);

运行结果