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Copy path22_Delete_Node_In_A_BST.java
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22_Delete_Node_In_A_BST.java
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Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.
Basically, the deletion can be divided into two stages:
Search for a node to remove.
If the node is found, delete the node.
Example 1:
Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.
Example 2:
Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.
Example 3:
Input: root = [], key = 0
Output: []
Constraints:
The number of nodes in the tree is in the range [0, 104].
-105 <= Node.val <= 105
Each node has a unique value.
root is a valid binary search tree.
-105 <= key <= 105
Follow up: Could you solve it with time complexity O(height of tree)?
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null) {
return null;
}
if (root.val < key) {
root.right = deleteNode(root.right, key);
}
else if (root.val > key) {
root.left = deleteNode(root.left, key);
}
else {
if (root.left == null) {
return root.right;
}
else if (root.right == null) {
return root.left;
}
TreeNode minimum = findMinimum(root.right);
root.val = minimum.val;
root.right = deleteNode(root.right, root.val);
}
return root;
}
private TreeNode findMinimum(TreeNode node) {
if (node == null) {
return null;
}
while (node.left != null) {
node = node.left;
}
return node;
}
}