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;
}
};

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