前言
思路及算法思维，指路 代码随想录。
题目来自 LeetCode。
day 22，难熬的周三~
题目详情
[235] 二叉搜索树的最近公共祖先
题目描述
解题思路
前提：二叉搜索树，且p、q均存在于二叉树上
思路：后序遍历，判断是否同处于某结点的左右子树上，因二叉搜索树有序性，可以判断结点在左子树还是右子树。
重点：有可能p、q为左右子树上，也有可能p为q的祖先。
代码实现
C语言
普通二叉树的最近公共祖先解法
c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode *traversal(struct TreeNode *root, struct TreeNode *p, struct TreeNode *q)
{
// 判空
if ((root == NULL) || (p == NULL) || (q == NULL))
{
return NULL;
}
// 判断该结点
if ((root == p) || (root == q))
{
return root;
}
// 左子树
struct TreeNode *leftNode = traversal(root->left, p, q);
// 右子树
struct TreeNode *rightNode = traversal(root->right, p, q);
// 判断是否找到最近公共祖先
if ((leftNode != NULL) &&(rightNode != NULL))
{
return root;
}
else if ((leftNode != NULL) &&(rightNode == NULL))
{
return leftNode;
}
else if ((leftNode == NULL) &&(rightNode != NULL))
{
return rightNode;
}
return NULL;
}
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
return traversal(root, p, q);
}二叉搜索树的有序性解法
c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode *traversal(struct TreeNode *root, struct TreeNode *p, struct TreeNode *q)
{
// 判空
if ((root == NULL) || (p == NULL) || (q == NULL))
{
return NULL;
}
// p、q结点均在左子树上
if ((root->val > p->val) && (root->val > q->val))
{
return traversal(root->left, p, q);
}
// p、q结点均在右子树上
if ((root->val < p->val) && (root->val < q->val))
{
return traversal(root->right, p, q);
}
// p、q一左一右或者一左一根或者一根一右，此时root均为最近公共祖先
return root;
}
struct TreeNode* lowestCommonAncestor(struct TreeNode* root, struct TreeNode* p, struct TreeNode* q) {
return traversal(root, p, q);
}[701] 二叉搜索树中的插入操作
题目描述
解题思路
前提：二叉搜索树
思路：由于二叉搜索树的有序性，插入的结点在对应叶子结点的左节点或右节点即可。
重点：二叉搜索树的有序性。
代码实现
C语言
递归
c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
void traversal(struct TreeNode **root, int val)
{
// 判空
if (*root == NULL)
{
*root = (struct TreeNode *)malloc(sizeof(struct TreeNode));
(*root)->val = val;
(*root)->left = NULL;
(*root)->right = NULL;
return ;
}
// 判断元素插入位置
if ((*root)->val > val)
{
traversal(&((*root)->left), val);
}
else if ((*root)->val < val)
{
traversal(&((*root)->right), val);
}
return ;
}
struct TreeNode* insertIntoBST(struct TreeNode* root, int val) {
traversal(&root, val);
return root;
}迭代
c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* insertIntoBST(struct TreeNode* root, int val) {
// 判空
if (root == NULL)
{
root = (struct TreeNode *)malloc(sizeof(struct TreeNode));
root->val = val;
root->left = NULL;
root->right = NULL;
return root;
}
// 原树非空
struct TreeNode *pre = NULL;
struct TreeNode *cur = root;
// 遍历至叶子结点位置
while (cur)
{
// 保存当前结点位置
pre = cur;
if (cur->val > val)
{
cur = cur->left;
}
else if (cur->val < val)
{
cur = cur->right;
}
}
// 插入节点
if (pre->val > val)
{
pre->left = (struct TreeNode *)malloc(sizeof(struct TreeNode));
pre->left->val = val;
pre->left->left = NULL;
pre->left->right = NULL;
}
else if (pre->val < val)
{
pre->right = (struct TreeNode *)malloc(sizeof(struct TreeNode));
pre->right->val = val;
pre->right->left = NULL;
pre->right->right = NULL;
}
return root;
}[450] 删除二叉搜索树中的节点
题目描述
解题思路
前提：二叉搜索树
思路：由于二叉搜索树的有序性，删除的结点比较容易遍历到，主要是处理该删除结点的左右子树的处理。
重点：该删除结点的左右子树的处理，如果左右子树有一为NULL时，则该子树代替删除结点的位置；如果左右子树均不为NULL时，则需要将左子树挂在右子树的最左侧叶子结点的左结点上，或者将右子树挂在左子树最右侧叶子结点的右节点上。
代码实现
C语言
二叉搜索树递归
c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* deleteNode(struct TreeNode* root, int key){
// 判空
if (root == NULL)
{
return root;
}
// 判断是否为删除结点
if (root->val == key)
{
// 如果该结点的左右子树至少一空
if (root->left == NULL)
{
return root->right;
}
else if (root->right == NULL)
{
return root->left;
}
else
{
// 寻找左侧子树的最右侧结点，放置该结点的右子树
struct TreeNode *pre = root;
struct TreeNode *cur = root->left;
while (cur)
{
pre = cur;
cur = cur->right;
}
pre->right = root->right;
root = root->left;
return root;
}
}
if (root->val > key)
{
root->left = deleteNode(root->left, key);
}
if (root->val < key)
{
root->right = deleteNode(root->right, key);
}
return root;
}二叉搜索树迭代
c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode *deleteOneNode(struct TreeNode *root)
{
// 判空
if (root == NULL)
{
return root;
}
// 删除结点
if (root->left == NULL)
{
return root->right;
}
if (root->right == NULL)
{
return root->left;
}
// 该结点两子树均不为空时，寻找左子树的最右叶子结点的右子树位置，指向原结点的右子树
struct TreeNode *cur = root->left;
while (cur->right)
{
cur = cur->right;
}
cur->right = root->right;
return root->left;
}
struct TreeNode* deleteNode(struct TreeNode* root, int key){
// 判空
if (root == NULL)
{
return root;
}
// 查找删除结点
struct TreeNode *pre = NULL;
struct TreeNode *cur = root;
while (cur)
{
if (cur->val == key)
{
break;
}
pre = cur;
if (cur->val > key)
{
cur = cur->left;
}
else if (cur->val < key)
{
cur = cur->right;
}
}
// 删除结点为根节点情况
if (pre == NULL)
{
return deleteOneNode(root);
}
// 判断删除结点为左子树，还是右子树
if ((pre->left) && (pre->left->val == key))
{
pre->left = deleteOneNode(pre->left);
}
if ((pre->right) && (pre->right->val == key))
{
pre->right = deleteOneNode(pre->right);
}
return root;
}普通二叉树删除方式：通过交换节点，然后删除叶子结点实现
c
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* deleteNode(struct TreeNode* root, int key){
// 判空
if (root == NULL)
{
return root;
}
// 判断是否为删除结点
if (root->val == key)
{
// 如果该结点的左右子树至少一空，此时删除该结点
if (root->left == NULL)
{
return root->right;
}
else if (root->right == NULL)
{
return root->left;
}
else
{
// 寻找该结点的左侧子树的最右侧结点，放置该结点的右子树,并交换两节点值
struct TreeNode *pre = root;
struct TreeNode *cur = root->left;
while (cur)
{
pre = cur;
cur = cur->right;
}
int tmp = root->val;
root->val = pre->val;
pre->val = tmp;
}
}
root->left = deleteNode(root->left, key);
root->right = deleteNode(root->right, key);
return root;
}今日收获
- 二叉搜索树的相关实现。