代码随想录算法训练营第二十天| 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

一、235 二叉搜索树的最近公共祖先

 

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

//迭代
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;
}

二、701 二叉搜索树中的插入操作

struct TreeNode* insertIntoBST(struct TreeNode* root, int val) {
    if(root==NULL){//找到要插入节点的位置
        struct TreeNode*node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
        node->val = val;
        //初始化新节点的左右节点
        node->left = NULL;
        node->right = NULL;
        return node;
    }
    if(val<root->val)
        root->left = insertIntoBST(root->left, val);
    if(val>root->val)
        root->right = insertIntoBST(root->right, val);
    return root;
}

三、450 删除二叉搜索树中的节点

struct TreeNode* deleteNode(struct TreeNode* root, int key) {
    if(root == NULL)//没找到
        return NULL;
    if(root->val == key){//找到后分4种情况
        if(root->left==NULL&&root->right == NULL)
            return NULL;
        else if(root->left!=NULL&&root->right==NULL)
            return root->left;
        else if(root->left == NULL&&root->right!=NULL)
            return root->right;
        else{//左不空右不空
            struct TreeNode*cur =  root->right;
            while(cur->left!=NULL){
                cur = cur->left;
            }
            cur->left = root->left;
            return root->right;
        }
    }
    if(key<root->val)
        root->left = deleteNode(root->left, key);
    if(key>root->val)
         root->right = deleteNode(root->right, key);
    return root;
}