(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;
}