450. Delete Node in a BST

最新推荐文章于 2019-02-11 20:06:35 发布

无名_1989

最新推荐文章于 2019-02-11 20:06:35 发布

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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。文章提供了两种实现方式，一种为递归实现，另一种为非递归实现。递归实现中详细解释了搜索并删除指定节点的过程，而非递归实现则通过遍历找到目标节点，并进行了相应的删除操作。

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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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
    
        if(!root) return NULL;
        if(root->val>key)
        {
            root->left = deleteNode(root->left, key);
            
        }else if(root->val<key)
        {
            root->right = deleteNode(root->right, key);
            
        }else
        {
            if(!root->left || !root->right)
            {
                root = (root->left) ? root->left : root->right;
                
            }else
            {
                TreeNode * cur = root->right;
                while(cur->left)
                {
                    cur = cur->left;
                }
                root->val = cur->val;
                root->right = deleteNode(root->right, cur->val);
            }
        }
        return root;
    }
};

非递归：class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        TreeNode *cur = root, *pre = NULL;
        while (cur) {
            if (cur->val == key) break;
            pre = cur;
            if (cur->val > key) cur = cur->left;
            else cur = cur->right;
        }
        if (!cur) return root;
        if (!pre) return del(cur);
        if (pre->left && pre->left->val == key) pre->left = del(cur);
        else pre->right = del(cur);
        return root;
    }
    TreeNode* del(TreeNode* node) {
        if (!node->left && !node->right) return NULL;
        if (!node->left || !node->right) {
            return (node->left) ? node->left : node->right;
        }
        TreeNode *pre = node, *cur = node->right;
        while (cur->left) {
            pre = cur;
            cur = cur->left;
        }
        node->val = cur->val;
        (pre == node ? node->right : pre->left) = cur->right;
        return node;
    }
};