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

Another valid answer is [5,2,6,null,4,null,7].

    5
   / \
  2   6
   \   \
    4   7

 

 

 解题思路：本题考查的主要是bst的查找和删除，难点主要在于树的拼接的处理，其中 root->right=deleteNode(root->right,root->val);简直不要太完美

/**
 * 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 root;
        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* temp=root->right;
            while(temp->left)temp=temp->left;
            root->val=temp->val;
            root->right=deleteNode(root->right,root->val);
        }
        return root;
    }
};