(BST二叉搜索树 1.2)Leetcode Delete no in a BST(删除二叉搜索树中的节点)

最新推荐文章于 2020-01-22 16:08:39 发布

原创最新推荐文章于 2020-01-22 16:08:39 发布 · 523 阅读

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本文介绍如何在二叉搜索树(BST)中删除指定键值的节点，并保持树的性质不变。文章详细阐述了搜索目标节点及删除操作的过程，包括特殊情况处理如删除节点有两子节点的情况。

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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:

Search for a node to remove.
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

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* deleteNode(struct TreeNode* root, int key) {
//如果节点为空,返回空
if(root == NULL){
return NULL;
}

//如果需要删除的节点 < 根节点的值
if(key < root->val){
//在左子树递归删除该节点
root->left = deleteNode(root->left,key);
}else if(key > root->val){//如果需要删除的节点 > 根节点的值
//在右子树递归删除该节点
root->right = deleteNode(root->right,key);
}else{//如果当前节点就是需要删除的节点

/*
如果是单边节点
*/
//如果左子树为空
if(root->left == NULL){
//直接删除根节点,将root直接指向右子树
root = root->right;
}else if(root->right == NULL){//如果右子树为空
//直接删除根节点,将root直接指向左子树
root = root->left;
}else{

/*
所有该节点拥有左右子树
*/
//寻找该节点的右子树的最右节点
struct TreeNode* cur = root->right;
while(cur->left){
cur = cur->left;
}
//将右子树的最左孩子节点的值付给根节点
root->val = cur->val;
//然后删除该节点
root->right = deleteNode(root->right,cur->val);
}
}

return root;
}