题目:

Given a root of Binary Search Tree with unique value for each node. Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal.

Given binary search tree:

    5
/ \
3 6
/ \
2 4

Remove 3, you can either return:

    5
/ \
2 6
\
4

or

    5
/ \
4 6
/
2

 

题解:

  这个题就是考察对二叉树操作的熟练程度,没有多少技巧,下面的程序中唯一能算作技巧的是更换node时有时并不需要切断其与左右节点和父节点的链接,只需要更换val值就可以了。

Solution 1 ()

class Solution {
public:
TreeNode* removeNode(TreeNode* root, int value) {
if (root == NULL)
return NULL;
TreeNode * head = new TreeNode();
head->left = root;
TreeNode * tmp = root, *father = head; while (tmp != NULL) {
if (tmp->val == value)
break;
father = tmp;
if (tmp->val > value)
tmp = tmp->left;
else
tmp = tmp->right;
}
if (tmp == NULL)
return head->left; if (tmp->right == NULL) {
if (father->left == tmp)
father->left = tmp->left;
else
father->right = tmp->left;
} else
if (tmp->right->left == NULL) {
if (father->left == tmp)
father->left = tmp->right;
else
father->right = tmp->right; tmp->right->left = tmp->left; } else {
father = tmp->right;
TreeNode * cur = tmp->right->left;
while (cur->left != NULL) {
father = cur;
cur = cur->left;
}
tmp->val = cur->val;
father->left = cur->right;
}
return head->left;
}
};

  用右子树最小值替代

Solution 2  ()

class Solution {
public:
TreeNode* removeNode(TreeNode* root, int value) {
if (root == nullptr) {
return nullptr;
}
if (root->val > value) {
root->left = removeNode(root->left, value);
} else if (root->val < value) {
root->right = removeNode(root->right, value);
} else {
// leaf node
if (root->left == nullptr && root->right == nullptr) {
root = nullptr;
} else if (root->left == nullptr) {
root = root->right;
} else if (root->right == nullptr) {
root = root->left;
} else {
TreeNode* tmp = findMin(root->right);
root->val = tmp->val;
root->right = removeNode(root->right, tmp->val);
}
} return root;
} TreeNode* findMin(TreeNode* root) {
while (root->left != nullptr) {
root = root->left;
}
return root;
}
};

  用左子树最大值替代

Solution 3 ()

class Solution {
public:
TreeNode* removeNode(TreeNode* root, int value) {
if (root == nullptr) {
return nullptr;
}
if (root->val > value) {
root->left = removeNode(root->left, value);
} else if (root->val < value) {
root->right = removeNode(root->right, value);
} else {
// leaf node
if (root->left == nullptr && root->right == nullptr) {
root = nullptr;
// only one child
} else if (root->left == nullptr) {
root = root->right;
} else if (root->right == nullptr) {
root = root->left;
// two child
} else {
TreeNode* tmp = findMax(root->left);
root->val = tmp->val;
root->left = removeNode(root->left, tmp->val);
}
} return root;
} TreeNode* findMax(TreeNode* root) {
while (root->right != nullptr) {
root = root->right;
}
return root;
}
};

【Lintcode】087.Remove Node in Binary Search Tree的更多相关文章

  1. Lintcode: Remove Node in Binary Search Tree

    iven a root of Binary Search Tree with unique value for each node. Remove the node with given value. ...

  2. Remove Node in Binary Search Tree 解答

    从BST中移除一个节点是比较复杂的问题,需要分好几种情况讨论. 如这篇文章,就讨论了删除节点 1.有无左右子树 2.只有右子树 3.只有左子树 三种情况. 一种简单些的思维是只考虑删除节点是否有右子树 ...

  3. 【leetcode】Convert Sorted List to Binary Search Tree

    Convert Sorted List to Binary Search Tree Given a singly linked list where elements are sorted in as ...

  4. 【PAT】1043 Is It a Binary Search Tree(25 分)

    1043 Is It a Binary Search Tree(25 分) A Binary Search Tree (BST) is recursively defined as a binary ...

  5. 【leetcode】701. Insert into a Binary Search Tree

    题目如下: Given the root node of a binary search tree (BST) and a value to be inserted into the tree, in ...

  6. 【LeetCode】501. Find Mode in Binary Search Tree 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...

  7. 【LeetCode】701. Insert into a Binary Search Tree 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.c ...

  8. 【leetcode】Convert Sorted Array to Binary Search Tree

    Convert Sorted Array to Binary Search Tree Given an array where elements are sorted in ascending ord ...

  9. 【树】Convert Sorted Array to Binary Search Tree

    题目: Given an array where elements are sorted in ascending order, convert it to a height balanced BST ...

随机推荐

  1. caffe-ubuntu1604-gtx850m-i7-4710hq----VGG_ILSVRC_16_layers.caffemodel

    c++调用vgg16: ./build/install/bin/classification \ /media/whale/wsWin10/wsCaffe/model-zoo/VGG16//deplo ...

  2. Iconfot阿里妈妈-css高级应用

    矢量图标替换教程 首先进入:http://www.iconfont.cn/ 搜索你分类的关键字---然后加入购物车 加入购物车之后,下载到本地用浏览器打开demo.html 把a class=“原来样 ...

  3. EntityFramework 6.0 修改一个已经存在的对象

    public void UpdateObj(someobject obj) { db.Entry(obj).State = EntityState.Modified; db.SaveChanges() ...

  4. mybatis的拦截器及分页机制

    https://blog.csdn.net/ssuperlg/article/details/79847889

  5. JavaWeb、J2-SE开发框架——Spring

    相关博客:   2.spring官网 1.使用Spring的jdbcTemplate进一步简化JDBC操作

  6. js关于变量作为if条件的真假问题

    var a = ""; if(a){ ..... }else{ .....} 以下情况会被认为返回false: "" 空的字符串 为 0 的数字 为 null ...

  7. /usr/bin/mysqld_safe_helper: Cannot change uid/gid (errno: 1) (转)

    From: https://www.rootusers.com/how-to-fix-mariadb-10-0-29-selinux-update-failure/ 安装mysql 10.0.29后, ...

  8. ssh无密码登陆屌丝指南

    [0]写在前面 由于ssh 实现的是免密码登陆,大致步骤是: 0.1) client通过ssh登陆到server: 0.2) server检查家目录下的.ssh文件, 并发送公钥文件 authoriz ...

  9. 奇妙的go语言(開始篇)

    [ 声明:版权全部.欢迎转载,请勿用于商业用途.  联系信箱:feixiaoxing @163.com] 从前接触脚本语言不多,可是自从遇到go之后,就開始慢慢喜欢上了这个脚本语言.go语言是goog ...

  10. iOS判断为空或者只为空格

    本文转载至 :http://www.cnblogs.com/superhappy/archive/2012/11/08/2761403.html 经常有需求 要判断不能为空,后台老是鄙视不做非空判断的 ...