C语言 | Leetcode C语言题解之第236题二叉树的最近公共祖先

题目：

题解：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
typedef struct road_t
{
    struct TreeNode *road_node; // 途径路径
    struct road_t *p_next;
}ROAD_T;
int road_fill(struct TreeNode* root, struct TreeNode* p, ROAD_T *road_node_now) // 路径填充
{
    ROAD_T *new_node = NULL;
    int flag = 0;
    if(!root)
    {
        return 0;
    }
    if(root == p) // 找到节点置起标志位，将途径路径存储
    {
        flag = 1;
    }
    flag |= road_fill(root->left, p, road_node_now);
    flag |= road_fill(root->right, p, road_node_now);
    if(flag)
    {
        new_node = (ROAD_T *)malloc(1 * sizeof(ROAD_T));
        if(!new_node)
        {
            printf("open space error");
            exit(0);
        }
        new_node->road_node = root;
        new_node->p_next = road_node_now->p_next;
        road_node_now->p_next = new_node;
        return flag;
    }
    return 0;
}
struct TreeNode *find_sncestor(ROAD_T *road_head_p, ROAD_T *road_head_q) // 寻找公共节点
{
    ROAD_T *now_p_node = road_head_p->p_next;
    ROAD_T *now_q_node = road_head_q->p_next;
    ROAD_T *last_same_node = NULL;
    while(now_p_node)
    {
        now_q_node = road_head_q;
        while(now_q_node)
        {
            if(now_p_node->road_node == now_q_node->road_node)
            {
                last_same_node = now_p_node->road_node;
            }
            now_q_node = now_q_node->p_next;
        }
        now_p_node = now_p_node->p_next;
    }
    return last_same_node;
}
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    ROAD_T *road_head_p = (ROAD_T *)calloc(1, sizeof(ROAD_T));
    ROAD_T *road_head_q = (ROAD_T *)calloc(1, sizeof(ROAD_T));
    if(!road_head_q || !road_head_p)
    {
        printf("open space error");
        exit(0);
    }
    road_fill(root, p, road_head_p); // 新建p节点路径
    road_fill(root, q, road_head_q); // 新建q节点路径
    return find_sncestor(road_head_p, road_head_q);
}