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The number of nodes in the tree is in the range [ 0 , 1 0 4 ] [0, 10^4] [0,104].
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− 1 0 5 < = N o d e . v a l < = 1 0 5 -10^5 <= Node.val <= 10^5 −105<=Node.val<=105
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Each node has a unique value.
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rootis a valid binary search tree. -
− 1 0 5 < = k e y < = 1 0 5 -10^5 <= key <= 10^5 −105<=key<=105
方法一：我写的。
-
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。
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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。
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根据目标节点的左右子树的有无指定删除方案：
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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）。
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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。
- 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。
- Recursively find the node that has the same value as the key, while setting the left/right nodes equal to the returned subtree.
- 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](https://docs.qq.com/doc/DSmRnRGxvUkxTREhO)
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;
}
/**
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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;
}
//方法二：别人写的，递归版，精简优雅了许多
public TreeNode deleteNode2(TreeNode root, int key) {
if(root == null){
return null;
}
if(key < root.val){
root.left = deleteNode2(root.left, key);
}else if(key > root.val){
root.right = deleteNode2(root.right, key);
}else{
if(root.left == null){
return root.right;
}else if(root.right == null){
return root.left;
}
TreeNode minNode = findMin(root.right);
root.val = minNode.val;
root.right = deleteNode2(root.right, root.val);
}
return root;
}
private TreeNode findMin(TreeNode node){
while(node.left != null){
node = node.left;
}
return node;
}
}
import static org.junit.Assert.*;
import org.junit.Test;
import com.lun.util.BinaryTree;
import com.lun.util.BinaryTree.TreeNode;
public class DeleteNodeInABSTTest {
@Test
public void test() {
DeleteNodeInABST obj = new DeleteNodeInABST();
TreeNode root1 = obj.deleteNode(BinaryTree.integers2BinaryTree(5,3,6,2,4,null,7), 3);
TreeNode expected1 = BinaryTree.integers2BinaryTree(5,4,6,2,null,null,7);
assertTrue(BinaryTree.equals(root1, expected1));
TreeNode root2 = obj.deleteNode(BinaryTree.integers2BinaryTree(5,3,6,2,4,null,7), 0);
TreeNode expected2 = BinaryTree.integers2BinaryTree(5,3,6,2,4,null,7);
assertTrue(BinaryTree.equals(root2, expected2));
assertNull(obj.deleteNode(null, 0));
//删除根节点
TreeNode root3 = obj.deleteNode(BinaryTree.integers2BinaryTree(5,3,6,2,4,null,7), 5);
TreeNode expected3 = BinaryTree.integers2BinaryTree(6,3,7,2,4);
assertTrue(BinaryTree.equals(root3, expected3));
assertNull(obj.deleteNode(new TreeNode(0), 0));
}
@Test
public void test2() {
DeleteNodeInABST obj = new DeleteNodeInABST();
TreeNode root1 = obj.deleteNode2(BinaryTree.integers2BinaryTree(5,3,6,2,4,null,7), 3);
TreeNode expected1 = BinaryTree.integers2BinaryTree(5,4,6,2,null,null,7);
assertTrue(BinaryTree.equals(root1, expected1));
TreeNode root2 = obj.deleteNode2(BinaryTree.integers2BinaryTree(5,3,6,2,4,null,7), 0);
TreeNode expected2 = BinaryTree.integers2BinaryTree(5,3,6,2,4,null,7);
assertTrue(BinaryTree.equals(root2, expected2));
assertNull(obj.deleteNode2(null, 0));
//删除根节点
TreeNode root3 = obj.deleteNode2(BinaryTree.integers2BinaryTree(5,3,6,2,4,null,7), 5);
TreeNode expected3 = BinaryTree.integers2BinaryTree(6,3,7,2,4);
assertTrue(BinaryTree.equals(root3, expected3));
最后
由于文档内容过多，为了避免影响到大家的阅读体验，在此只以截图展示部分内容
