# 450. Delete Node in a BST

題目：<https://leetcode.com/problems/delete-node-in-a-bst/>

解法：找出 key 節點，如果 key 節點是葉節點，則刪除它，反之，找出左子樹的最大節點或是右子樹的最小節點，將 key 節點的值用此節點值取代後，再刪除此節點。

Python3：

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

def listToTree(arr):
    nodes = [TreeNode(arr[0])]
    i = 0
    j = 1
    length = len(arr)
    while j < length:
        if arr[j] is not None:
            tmp = TreeNode(arr[j])
            nodes[i].left = tmp
            nodes.append(tmp)
        j += 1
        if j >= length:
            break
        if arr[j] is not None:
            tmp = TreeNode(arr[j])
            nodes[i].right = tmp
            nodes.append(tmp)
        j += 1
        i += 1
    return nodes[0]

def treeToList(root):
    if not root:
        return []
    from collections import deque
    nodes = deque()
    nodes.append(root)
    arr = []
    arr.append(root.val)
    while len(nodes):
        node = nodes.popleft()
        if node.left:
            nodes.append(node.left)
            arr.append(node.left.val)
        else:
            arr.append(None)
        if node.right:
            nodes.append(node.right)
            arr.append(node.right.val)
        else:
            arr.append(None)
    return arr

class Solution:
    def deleteNode(self, root: TreeNode, key: int) -> TreeNode:
        if not root:
            return None
        delParent = None
        keyNode = root
        while keyNode:
            if key < keyNode.val:
                delParent = keyNode
                keyNode = keyNode.left
                delDir = 0
            elif key > keyNode.val:
                delParent = keyNode
                keyNode = keyNode.right
                delDir = 1
            else:
                break
        if not keyNode:
            return root
        if keyNode.left:
            delParent = keyNode
            cur = keyNode.left
            delDir = 0
            while cur.right:
                delParent = cur
                cur = cur.right
                delDir = 1
            keyNode.val = cur.val
        elif keyNode.right:
            delParent = keyNode
            cur = keyNode.right
            delDir = 1
            while cur.left:
                delParent = cur
                cur = cur.left
                delDir = 0
            keyNode.val = cur.val
        else:
            cur = keyNode
        if delParent:
            if delDir == 0:
                delParent.left = cur.left if cur.left else cur.right
            elif delDir == 1:
                delParent.right = cur.left if cur.left else cur.right
            return root
        else:
            return None
        
if __name__ == '__main__':
    arr = [4,None,7,6,8,5,None,None,9]
    root = listToTree(arr)
    print(treeToList(root))
    val = 7
    
    ans = Solution().deleteNode(root, val)
    print(treeToList(ans))

C：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
};

struct TreeNode* strToTree(char *str) {
    char *token = strtok(str, ",");
    struct TreeNode **nodes = (struct TreeNode**)malloc(sizeof(struct TreeNode*));
    nodes[0] = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    nodes[0]->val = atoi(token);
    nodes[0]->left = NULL;
    nodes[0]->right = NULL;
    int i = 0, len = 1;
    token = strtok(NULL, ",");
    while (token != NULL) {
        if (strcmp(token, "null") != 0) {
            len++;
            nodes = (struct TreeNode**)realloc(nodes, sizeof(struct TreeNode*) * len);
            nodes[len - 1] = (struct TreeNode*)malloc(sizeof(struct TreeNode));
            nodes[len - 1]->val = atoi(token);
            nodes[len - 1]->left = NULL;
            nodes[len - 1]->right = NULL;
            nodes[i]->left = nodes[len - 1];
        }
        token = strtok(NULL, ",");
        if (token == NULL)
            break;
        if (strcmp(token, "null") != 0) {
            len++;
            nodes = (struct TreeNode**)realloc(nodes, sizeof(struct TreeNode*) * len);
            nodes[len - 1] = (struct TreeNode*)malloc(sizeof(struct TreeNode));
            nodes[len - 1]->val = atoi(token);
            nodes[len - 1]->left = NULL;
            nodes[len - 1]->right = NULL;
            nodes[i]->right = nodes[len - 1];
        }
        token = strtok(NULL, ",");
        i++;
    }
    return nodes[0];
}

void printLevelOrderWithNull(struct TreeNode* root) {
    struct TreeNode * bfs[10000] = {};
    bfs[0] = root;
    int i = 0, len = 1;
    printf("%d ", root->val);
    while (i < len) {
        if (bfs[i]->left != NULL) {
            bfs[len++] = bfs[i]->left;
            printf("%d ", bfs[i]->left->val);
        } else
            printf("null ");
        if (bfs[i]->right != NULL) {
            bfs[len++] = bfs[i]->right;
            printf("%d ", bfs[i]->right->val);
        } else
            printf("null ");
        i++;
    }
    printf("\n");
}

struct TreeNode* deleteNode(struct TreeNode* root, int key) {
    if (root == NULL)
        return NULL;
    struct TreeNode *delParent = NULL, *keyNode = root, *cur;
    int delDir = -1;
    while (keyNode != NULL) {
        if (key < keyNode->val) {
            delParent = keyNode;
            keyNode = keyNode->left;
            delDir = 0;
        } else if (key > keyNode->val) {
            delParent = keyNode;
            keyNode = keyNode->right;
            delDir = 1;
        } else
            break;
    }
    if (keyNode == NULL)
        return root;
    if (keyNode->left) {
        delParent = keyNode;
        cur = keyNode->left;
        delDir = 0;
        while (cur->right) {
            delParent = cur;
            cur = cur->right;
            delDir = 1;
        }
    } else if (keyNode->right) {
        delParent = keyNode;
        cur = keyNode->right;
        delDir = 1;
        while (cur->left) {
            delParent = cur;
            cur = cur->left;
            delDir = 0;
        }
    } else
        cur = keyNode;
    keyNode->val = cur->val;
    
    if (delParent == NULL)
        return NULL;
    if (delDir == 0)
        delParent->left = cur->left != NULL ? cur->left : cur->right;
    else if (delDir == 1)
        delParent->right = cur->left != NULL ? cur->left : cur->right;
    return root;
}

int main()
{
    char input[] = "4,null,7,6,8,5,null,null,9";
    struct TreeNode* root = strToTree(input);
    printLevelOrderWithNull(root);
    printf("\n");
    int val = 7;
    
    struct TreeNode* ans = deleteNode(root, val);
    printLevelOrderWithNull(ans);
    return 0;
}

###### tags: `leetcode` `binary tree` `binary search tree`