450. Delete Node in a BST

二叉搜索树删除节点

最新推荐文章于 2020-06-10 19:21:07 发布

zibingling777

最新推荐文章于 2020-06-10 19:21:07 发布

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分类专栏：刷题笔记文章标签： Tree

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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。文章首先概述了问题，随后给出了详细的解决方案，包括当根节点为待删节点时的四种情况处理方式，并通过递归方式更新左右子树。

1. Description

Given a BST and a key, delete the node with the given key in the BST.

2. Solution

If the root's value is equal to k, there are four cases to analyse.

Otherwise, try to delete the node with k from the left-child and right-child, update the root's two child, and return the root.

3. Code

    TreeNode* deleteNode(TreeNode* root, int key) {
        if(!root) return NULL;
        if(root->val ==key){
            TreeNode* l = root->left;
            TreeNode* r = root->right;
            if(!l&&!r) return NULL;
            else if(!r) return l;
            else if(!l) return r;
            else{
                root = r;
                while(r->left){
                    r=r->left;
                }
                r->left = l;
                return root;
            }
        }
        else{
            TreeNode* l = deleteNode(root->left,key);
            TreeNode* r = deleteNode(root->right,key);
            root->left = l;
            root->right = r;
            return root;            
        }
    }