(BST二叉搜索树 1.2)Leetcode Delete no in a BST(删除二叉搜索树中的节点)

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

/**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     struct TreeNode *left;
  *     struct TreeNode *right;
  * };
  */
 struct TreeNode* deleteNode(struct TreeNode* root, int key) {
     //如果节点为空,返回空
     if(root == NULL){
         return NULL;
     }
     
     //如果需要删除的节点 < 根节点的值
     if(key < root->val){
         //在左子树递归删除该节点
         root->left = deleteNode(root->left,key);
     }else if(key > root->val){//如果需要删除的节点 > 根节点的值
         //在右子树递归删除该节点
         root->right = deleteNode(root->right,key);
     }else{//如果当前节点就是需要删除的节点
         
         /*
         如果是单边节点
         */
         //如果左子树为空
         if(root->left == NULL){
             //直接删除根节点,将root直接指向右子树
             root = root->right;
         }else if(root->right == NULL){//如果右子树为空
             //直接删除根节点,将root直接指向左子树
             root = root->left;
         }else{
             
             /*
             所有该节点拥有左右子树
             */
             //寻找该节点的右子树的最右节点
             struct TreeNode* cur = root->right;
             while(cur->left){
                 cur = cur->left;
             }
             //将右子树的最左孩子节点的值付给根节点
             root->val = cur->val;
             //然后删除该节点
             root->right = deleteNode(root->right,cur->val);
         }
     }
     
     return root;
 }