LintCode Remove Node in Binary Search Tree

Given a root of Binary Search Tree with unique value for each node. Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal.

以后估计不会再写了

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
class Solution {
public:
    /**
     * @param root: The root of the binary search tree.
     * @param value: Remove the node with given value.
     * @return: The root of the binary search tree after removal.
     */
    TreeNode* removeNode(TreeNode* root, int value) {
        // write your code here
        if (root == NULL) {
            return NULL;
        }

        if (root->val > value) {
            root->left = removeNode(root->left, value);
            return root;
        } else if (root->val < value) {
            root->right= removeNode(root->right, value);
            return root;
        }
        if (root->left == NULL && root->right == NULL) {
            delete root;
            return NULL;
        }
        if (root->left == NULL || root->right == NULL) {
            TreeNode* child = root->left ? root->left : root->right;
            root->left = child->left;
            root->right = child->right;
            root->val = child->val;
            delete child;
            return root;
        }
        TreeNode* min2move = findMin(root->right);
        root->val = min2move->val;
        root->right = removeNode(root->right, min2move->val);
        return root;
    }
    
    TreeNode* findMin(TreeNode* root) {
        TreeNode* curr = root;
        TreeNode* prev = NULL;
        while (curr != NULL) {
            prev = curr;
            curr = curr->left;
        }
        return prev;
    }
};
原文地址:https://www.cnblogs.com/lailailai/p/4832835.html