450. Delete Node in a BST
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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.
Note: Time complexity should be O(height of tree).
Example:
root = [5,3,6,2,4,null,7] key = 3 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. 5 / 4 6 / 2 7 Another valid answer is [5,2,6,null,4,null,7]. 5 / 2 6 4 7
class Solution { public TreeNode deleteNode(TreeNode root, int key) { if(root == null) { return root; } //find the target if(root.val < key) { root.right = deleteNode(root.right, key); return root; } else if(root.val > key) { root.left = deleteNode(root.left, key); return root; } //No child or only one child if(root.left == null && root.right == null) { return null; } else if(root.left == null) { return root.right; } else if(root.right == null) { return root.left; } //Two children if(root.right.left == null) { root.right.left = root.left; return root.right; } else { TreeNode smallest = deleteSmallest(root.right); smallest.left = root.left; smallest.right = root.right; return smallest; } } private TreeNode deleteSmallest(TreeNode root) { TreeNode cur = root.left; TreeNode pre = root; while(cur.left != null) { pre = cur; cur = cur.left; } pre.left = cur.right; return cur; } }
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**Leetcode 450. Delete Node in a BST