代码随想录第22天|235. 二叉搜索树的最近公共祖先 , 701.二叉搜索树中的插入操作,450.删除二叉搜索树中的节点

LeetCode235.二叉搜索树的最近公共祖先

题目链接:235. 二叉搜索树的最近公共祖先 - 力扣(LeetCode)

思路:

搜索一条边的写法:

if (递归函数(root->left)) return ;
if (递归函数(root->right)) return ;

搜索整个树写法:

left = 递归函数(root->left);
right = 递归函数(root->right);
left与right的逻辑处理;
/**
 * 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) return root;
        if(root->val > p->val && root->val > q->val) {
            TreeNode* left = lowestCommonAncestor(root->left,p,q);
            if(left != NULL) return left;
        }
        else if(root->val < p->val && root->val < q->val) {
            TreeNode* right = lowestCommonAncestor(root->right,p,q);
            if(right != NULL) return right;
        }
        else return root;
        return NULL;
    }
};

LeetCode701.二叉搜索树中的插入操作

题目链接:701. 二叉搜索树中的插入操作 - 力扣(LeetCode)

思路:

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

LeetCode450.删除二叉搜索树中的节点

题目链接:450. 删除二叉搜索树中的节点 - 力扣(LeetCode)

思路:

/**
 * Definition for a binary tree node.
 * 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* deleteNode(TreeNode* root, int key) {
        if(root == NULL) return root;//第一种情况,没找到节点
        if(root->val == key) {
            if(root->left == NULL && root->right == NULL) {//第二种,叶子节点,左右都为空
                //记得每删除一个释放内存,如:delete root;
                return NULL;
            }
            else if(root->left != NULL && root->right == NULL) {//第三种,左孩子不为空,右孩子为空
                return root->left;
            }
            else if(root->left == NULL && root->right != NULL) {//第四种,左孩子为空,右孩子不为空
                return root->right;
            }
            else {//第五种,左右都不为空
                TreeNode* cur = root->right;
                while(cur->left != NULL) {
                    cur = cur->left;
                }
                cur->left = root->left; //要删除的节点左子树放在cur左孩子位置上
                root = root->right;//返回旧root的右孩子为root新节点
                return root;
            }
        }
        if(root->val > key) {
            root->left = deleteNode(root->left,key);
        }
        if(root->val < key) {
            root->right = deleteNode(root->right,key);
        }
        return root;
    }
};

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