lintcode87. 删除二叉查找树的节点 leetcode 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

思路对

自己的写法没调出了

放弃看题解了

选左子树最右边点和右子树最左边的点都可以

/**
 * 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) {
        TreeNode* current = root;
        if (current == NULL) return current;
        if (current -> val > key)
            current->left =  deleteNode(root->left, key);
        else if (current -> val < key)
            current->right =  deleteNode(root->right, key);
        else {
            if (!root->left || !root->right) {
                root = root->left?root->left:root->right;
            } else {
                TreeNode * tmp = current->left;
                while(tmp->right) tmp = tmp->right;
                current->val = tmp->val;
                root->left = deleteNode(root->left, tmp->val);
            }
        }
        return root;
    }
};