450. Delete Node in a BST

最新推荐文章于 2019-02-11 20:06:35 发布

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最新推荐文章于 2019-02-11 20:06:35 发布

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分类专栏： leetcode

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本文介绍了一种从二叉搜索树(BST)中删除指定键的方法。文章详细阐述了如何通过递归查找目标节点，并提供了两种有效更新树结构的方式：一种是将左子树直接挂载到右子树最小节点的左侧；另一种是将右子树挂载到左子树最大节点的右侧。

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

思路：递归找到要删除的节点root，然后在右子树root.right中找一个最小的node（也可以在左子树找最大的）minNode，然后把root.left放到minNode.left

package l450;
public class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null)	return root;
        
        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;
        	if(root.right == null)	return root.left;
        	
        	// find right smallest
        	TreeNode rigthSmallest = root.right;
        	while(rigthSmallest.left != null)	rigthSmallest = rigthSmallest.left;
        	
        	// change structure of tree
        	rigthSmallest.left = root.left;
        	return root.right;
        }
        
        return root;
    }
}