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).
一刷
题解：
这题用recursive做最适合。因为如果要删除的点同时拥有左子树和右子树，然而又要删去right子树中的leftmost
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if(root == null) return null;
if(key<root.val) root.left = deleteNode(root.left, key);
else if(key > root.val) root.right = deleteNode(root.right, key);
else{
if(root.left == null) return root.right;
else if(root.right == null) return root.left;
TreeNode minNode = findMin(root.right);
root.val = minNode.val;
root.right = deleteNode(root.right, root.val);
}
return root;
}
private TreeNode findMin(TreeNode root){
while(root.left!=null) root = root.left;
return root;
}
}
