## 题目描述：

LeetCode 450. Delete Node in a BST

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

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

5
/ \
2   6
\   \
4   7
```

## 题目大意：

1. 寻找待删除节点。
2. 如果节点找到，删掉这个节点。

## Python代码：

``````# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def deleteNode(self, root, key):
"""
:type root: TreeNode
:type key: int
:rtype: TreeNode
"""
pre, cur = None, root
while cur and cur.val != key:
pre = cur
if key < cur.val:
cur = cur.left
elif key > cur.val:
cur = cur.right
if not cur: return root

ncur = cur.right
if cur.left:
ncur = cur.left
self.maxChild(cur.left).right = cur.right

if not pre: return ncur

if pre.left == cur:
pre.left = ncur
else:
pre.right = ncur
return root

def maxChild(self, root):
while root.right:
root = root.right
return root
``````

## Java代码：

``````/**
* 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;
}
root.val = findMax(root.left).val;
root.left = deleteNode(root.left, root.val);
}
return root;
}

private TreeNode findMax(TreeNode node){
while(node.right != null){
node = node.right;
}
return node;
}
}
``````

Pingbacks已关闭。