450. Delete Node in a BST ---------M

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.

Follow up: Can you solve it with time complexity O(height of tree)?

 

Example 1:

Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: 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 above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

Example 2:

Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.

Example 3:

Input: root = [], key = 0
Output: []

 

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -105 <= Node.val <= 105
• Each node has a unique value.
• root is a valid binary search tree.
• -105 <= key <= 105

 A:

**************递归实现********************

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        TreeNode* parent = new TreeNode(0);
        parent->left = root;
        helper(root, parent, true, key);
        return parent->left;
    }

private:
    void helper(TreeNode* root, TreeNode* parent, bool isRootLeftChild, int key) {
        if (!root)
            return;
        if (root->val == key) { // copy node, and remove the most
            if (root->left) {   // if left has value. copy its right most value,
                                // and then remove that value
                auto runner = root->left;
                while (runner->right) {
                    runner = runner->right;
                }
                root->val = runner->val;
                helper(root->left, root, true,
                       root->val); // continue to remvoe the copied value
            } else if (root->right) {
                auto runner = root->right;
                while (runner->left) {
                    runner = runner->left;
                }
                root->val = runner->val;
                helper(root->right, root, false, root->val);
            } else { // no child, we can simply delete this node
                if (isRootLeftChild) {
                    parent->left = nullptr;
                } else {
                    parent->right = nullptr;
                }
                delete(root);
            }
        } else if (root->val > key) {
            helper(root->left, root, true, key);
        } else {
            helper(root->right, root, false, key);
        }
    }
};

********************   迭代。   循环就是个实现题。考的是细心 ********

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        TreeNode header;
        header.left = root;
        TreeNode * pre = &header;
        TreeNode * cur = root;
        bool preLeft = true;
        while(cur!= nullptr && cur->val != key){
            if(cur->val <key){
                preLeft = false;
                pre = cur;
                cur = cur->right;
            }else{
                preLeft = true;
                pre = cur;
                cur = cur->left;
            }
        }
        if(cur!= nullptr){
            if(cur->left== nullptr){
                if(preLeft)
                    pre->left = cur->right;
                else
                    pre->right = cur->right;
            }else if(cur->right == nullptr){                
                if(preLeft)
                    pre->left = cur->left;
                else
                    pre->right = cur->left;
            }else{ // have two children
                // find the biggest on left children, and detach
                auto p = cur;
                auto child = cur->left;
                while(child->right){
                    p = child;
                    child = child->right;
                }
                cur->val = child->val;
                if(p==cur){
                    p->left = child->left;  // ERROR made, had assigned nullptr
                }else{
                    p->right = child->left;
                }
            }
        }
        return header.left;
    }    
};

ERROR：

就是p->left的时候，以为到了最后一个了，child节点的左右都是Null了， 然而不是的。哎