235. 二叉搜索树的最近公共祖先

题目链接

解: 迭代

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    while (root != NULL) {
        if (root->val > p->val && root->val > q->val) {
            root = root->left;
        } else if (root->val < p->val && root->val < q->val) {
            root = root->right;
        } else return root;
    }

    return NULL;
}

解: 递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode *travel(struct TreeNode *node, struct TreeNode *p, struct TreeNode *q) {
    if (node == NULL) return NULL;

    if (node->val > p->val && node->val > q->val) {
        struct TreeNode *left = travel(node->left, p, q);
        if (left != NULL) return left;
    }

    if (node->val < p->val && node->val < q->val) {
        struct TreeNode *right = travel(node->right, p, q);
        if (right != NULL) return right;
    }

    return node;
}

struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
    return travel(root, p, q);
}

701. 二叉搜索树中的插入操作

题目链接

解: 递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* insertIntoBST(struct TreeNode* root, int val) {
    if (root == NULL) {
        struct TreeNode *node = (struct TreeNode *)malloc(sizeof(*node));
        node->val = val;
        node->left = node->right = NULL;
        return node;
    }

    if (root->val > val) root->left = insertIntoBST(root->left, val);
    if (root->val < val) root->right = insertIntoBST(root->right, val);

    return root;
}

解: 迭代

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
struct TreeNode* insertIntoBST(struct TreeNode* root, int val) {
    struct TreeNode *node = (struct TreeNode *)malloc(sizeof(*node));
    node->val = val;
    node->left = node->right = NULL;

    if (root == NULL) {
        return node;
    }
    
    struct TreeNode *pre = NULL, *cur = root;

    while (cur != NULL) {
        pre = cur;
        if (cur->val > val) cur = cur->left;
        else cur = cur->right;
    }

    if (pre->val > val) {
        pre->left = node;
    } else {
        pre->right = node;
    }

    return root;
}

450. 删除二叉搜索树中的节点

题目链接

解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */


struct TreeNode* deleteNode(struct TreeNode* root, int key){
    if (root == NULL) return NULL;

    if (root->val == key) {
        if (root->left == NULL && root->right == NULL) {
            free(root);
            return NULL;
        } else if (root->left != NULL && root->right == NULL) {
            struct TreeNode *node = (struct TreeNode *)malloc(sizeof(*node));
            node = root->left;
            free(root);
            return node;
        } else if (root->left == NULL && root->right != NULL) {
            struct TreeNode *node = (struct TreeNode *)malloc(sizeof(*node));
            node = root->right;
            free(root);
            return node;
        } else {
            struct TreeNode *cur = root->right;
            while (cur->left != NULL) {
                cur = cur->left;
            }
            cur->left = root->left;

            struct TreeNode *tmp = root;
            root = root->right;
            free(tmp);
            return root;
        }
    }

    if (root->val > key) root->left = deleteNode(root->left, key);
    if (root->val < key) root->right = deleteNode(root->right, key);
    
    return root;
}

解: 迭代

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode *delete_one_node(struct TreeNode *node) {
    if (node->right == NULL) return node->left;

    // 返回右节点的情况合并
    struct TreeNode *cur = node->right;
    while (cur->left != NULL) {
        cur = cur->left;
    }

    cur->left = node->left;
    return node->right;
}


struct TreeNode* deleteNode(struct TreeNode* root, int val){
    if (root == NULL) return NULL;

    struct TreeNode *cur = root, *pre = NULL;

    while (cur != NULL) {
        if (cur->val == val) break;
        pre = cur;
        if (cur->val > val) cur = cur->left;
        else cur = cur->right;
    } 

    if (pre == NULL) {
        return delete_one_node(cur);
    }

    if (pre->left != NULL && pre->left->val == val) {
        pre->left = delete_one_node(cur);
    }

    if (pre->right != NULL && pre->right->val == val) {
        pre->right = delete_one_node(cur);
    }

    return root;
}