LeetCode - Medium - 450

Example 1:

Input: root = [5,3,6,2,4,null 《大厂前端面试题解析+Web核心总结学习笔记+企业项目实战源码+最新高清讲解视频》无偿开源 徽信搜索公众号【编程进阶路】 ,7], key = 3

Output: [5,4,6,2,null,null,7]

Explanation: 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 above BST.

Please notice that another valid answer is [5,2,6,null,4,null,7] and it’s also accepted.

Example 2:

Input: root = [5,3,6,2,4,null,7], key = 0

Output: [5,3,6,2,4,null,7]

Explanation: The tree does not contain a node with value = 0.

Example 3:

Input: root = [], key = 0

Output: []

Constraints:

• The number of nodes in the tree is in the range [ 0 , 1 0 4 ] [0, 10^4] [0,104].

• − 1 0 5 < = N o d e . v a l < = 1 0 5 -10^5 <= Node.val <= 10^5 −105<=Node.val<=105

• Each node has a unique value.

• root is a valid binary search tree.

• − 1 0 5 < = k e y < = 1 0 5 -10^5 <= key <= 10^5 −105<=key<=105

[](()Analysis

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方法一：我写的。

1. 找出目标节点，找到即返回它和它的父节点，找不到直接结束算法。

2. 如果目标节点是根节点，就给它弄个临时父节点，原根节点成为它的左子树。

3. 根据目标节点的左右子树的有无指定删除方案：

• 目标节点是叶子节点，或只有一个左子树，或只有一个右子树，则采用类似单链表删除节点方法（这就需要用到目标节点的父节点）。

• 目标节点都有左右子树，则从右子树中查找出最小值节点，让它值替换目标节点值，随后在右子树移除最小值节点。由于BST的性质，最小值节点有么是叶子节点，有么只有右子树，只有这两种情况，则放心采用类似单链表删除节点方法移除最小值节点。

4. 最后，返回根节点。如果目标节点是根节点，则返回临时父节点的左子树。

方法二：别人写的，递归版，精简优雅。

1. Recursively find the node that has the same value as the key, while setting the left/right nodes equal to the returned subtree.

2. Once the node is found, have to handle the below 4 cases:

*   node doesn’t have left or right - return null

*   node only has left subtree- return the left subtree

*   node only has right subtree- return the right subtree

*   node has both left and right - find the minimum value in the right subtree, set that value to the currently found node, then  

    recursively delete the minimum value in the right subtree  

    [link](()

[](()Submission

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import com.lun.util.BinaryTree.TreeNode;

public class DeleteNodeInABST {

//方法一：我写的

public TreeNode deleteNode(TreeNode root, int key) {

TreeNode[] result = find(root, key);

if(result != null) {

TreeNode parent = result[0], target = result[1];

TreeNode fakeRoot = null;

if(parent == null) {//删除根节点，弄个假根节点，原

fakeRoot = new TreeNode(Integer.MAX_VALUE, root, null);

}

if(target.left != null && target.right != null) {//既有左子树，有右子树

TreeNode replaceOne = findReplaceOneFromRight(target, target.right);

target.val = replaceOne.val;

}else {

removeEasyOne(parent == null ? fakeRoot: parent, target);

}

if(parent == null) {//删除根节点的情况

root = fakeRoot.left;

fakeRoot.left = null;

}

}

return root;

}

private TreeNode[] find(TreeNode root, int key) {

TreeNode last = null, curr = root;

while(curr != null) {

if(key < curr.val) {

last = curr;

curr = curr.left;

}else if(curr.val < key){

last = curr;

curr = curr.right;

}else {

return new TreeNode[] {last, curr};

}

}

return null;

}

/**

• target是叶子节点，target只有左子树，target只有右子数，这三种情况可以用链表式方法的删除节点

*/

private void removeEasyOne(TreeNode parent, TreeNode target) {

if(target.left == null && target.right == null) {//target是叶子节点

if(parent.left == target){

parent.left = null;

}else {

parent.right = null;

}

}else if(target.left != null && target.right == null){//target只有左子树

if(parent.left == target){

parent.left = target.left;

target.left = null;

}else {

parent.right = target.left;

target.left = null;

}

}else if(target.left == null && target.right != null){//target只有右子数

if(parent.left == target){

parent.left = target.right;

target.right = null;

}else {

parent.right = target.right;

target.right = null;

}

}

}

//查找右子树的最小值

private TreeNode findReplaceOneFromRight(TreeNode last, TreeNode child) {

if(child == null) return null;

while(child.left != null) {//

last = child;

child = child.left;

}

removeEasyOne(last, child);

return child;

}