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:

解析

刪除給定值得節點

解法1：遞迴

首先判斷根節點是否為空，由於BST的性質，可以遞迴定位到要刪除的節點。當前節點不等於key時，根據大小關係遞迴呼叫左右節點。
當前節點就是要刪除的節點時，首先判斷是否有一個子節點不存在，將root指向存在的子節點；當左右子節點都存在時，在右子樹中找到最小值得節點，並將其值賦給當前節點，再遞迴呼叫右子樹刪除其最小節點。

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if(!root) return NULL;
        if(root->val > key)
            root->left = deleteNode(root->left, key);
        else if(root->val < key)
            root->right = deleteNode(root->right, key);
        else{
            if(!root->left || !root->right)
                root = root->left ? root->left:root->right;
            else{
                TreeNode* cur = root->right;
                while(cur->left) cur = cur->left;
                root->val = cur->val;
                root->right = deleteNode(root->right, cur->val);
            }
        }
        return root;
    }
};
 

解法2：迭代

首先構建一個父節點pre以及當前節點cur，找到key的節點位置為cur，如果cur為空返回root，如果pre為空說明刪除節點是根節點，呼叫del函式刪除cur，如果刪除節點是pre->left，呼叫del刪除cur，並且pre->left=del(cur)，反之，pre->right=del(cur)。
下面解釋刪除函式del(node)。
刪除函式是為了刪除當前節點，所以當node為葉子結點時，返回NULL;
當node有一個子節點時，返回那個子節點；
當node有兩個子節點時，先在右子樹中找到最小值節點cur, 並記錄其父節點pre，並賦值給node，當pre為node時，node->right = cur->right，反之，pre->right = cur->right；返回node。

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        TreeNode* cur = root, *pre = NULL;
        while(cur){
            if(cur->val == key) break;
            pre = cur;
            if(cur->val > key)
                cur = cur->left;
            else
                cur = cur->right;
        }
        if(!cur) return root;
        if(!pre) return del(cur);
        if(pre->left && pre->left->val == key) 
            pre->left = del(cur);
        else
            pre->right = del(cur);
        return root;
    }
    
    TreeNode* del(TreeNode* node){
        if(!node->left && !node->right) return NULL;
        if(!node->left || !node->right)  
            return node->left ? node->left : node->right;
        TreeNode* pre = node, *cur = node->right;
        while(cur->left){
            pre = cur;
            cur = cur->left;
        }
        node->val = cur->val;
        if(pre == node)
            node->right = cur->right;
        else
            pre->left = cur->right;
        return node;
    }
};

解法3：通用二叉樹

遍歷所有節點，然後刪除值為key的節點

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return NULL;
        if (root->val == key) {
            if (!root->right) return root->left;
            else {
                TreeNode *cur = root->right;
                while (cur->left) cur = cur->left;
                swap(root->val, cur->val);
            }
        }
        root->left = deleteNode(root->left, key);
        root->right = deleteNode(root->right, key);
        return root;
    }
};

參考

http://www.cnblogs.com/grandyang/p/6228252.html