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[leetcode] problem 450 – delete node in a bst

admin 11月 13, 2020 0

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:

  1. Search for a node to remove.
  2. If the node is found, delete the node.

Note

Time complexity should be O(height of tree).

Example

root = [5,3,6,2,4,null,7]
key = 3

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5
 
    5
   / 
  3   6
 /    
2   4   7

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 following BST.

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5
 
    5
   / 
  4   6
 /     
2       7

Another valid answer is [5,2,6,null,4,null,7].

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5
 
  5
 / 
2   6
    
  4   7

Code

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public class  {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}
 
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public TreeNode deleteNode(TreeNode root, int key) {
    if (root == null)
        return null;

    if (root.val > key)
        root.left = deleteNode(root.left, key);
    else if (root.val < key)
        root.right = deleteNode(root.right, key);
    else {
        if (root.left == null || root.right == null)
            root = (root.left != null) ? root.left : root.right;
        else {
            TreeNode node = root.right;

            while (node.left != null)
                node = node.left;

            root.val = node.val;
            root.right = deleteNode(root.right, node.val);
        }
    }

    return root;
}
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