算法day17-二叉树（7）
目录[*]二叉搜索树的最近公共祖先
[*]二叉搜索树中的插入操作
[*]删除二叉搜索树中的节点
一、二叉搜索树的最近公共祖先
https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-search-tree/description/?envType=problem-list-v2&envId=8At1GmaZ
方法一：
class Solution {
TreeNode res = new TreeNode(-1);
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root == null){
return null;
}
// System.out.print(res == null);
int x = p.val, y = q.val;
if(x > y){
int temp = x;
x = y;
y = temp;
}
findAncestor(root, x, y);
return res;
}
public void findAncestor(TreeNode root, int x, int y){
if(res.val != -1){
return;
}
if(root.val >= x && root.val <= y){
//返回答案
// System.out.println("find" + " " +root.val);
res = root;
return;
}
if(root.val > x && root.val > y ){
findAncestor(root.left, x, y);
}else if(root.val < x && root.val < y){
findAncestor(root.right, x, y);
}
}
} 方法二：非递归。
class Solution{
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
while((long)(root.val - p.val) * (root.val - q.val) > 0){
root = p.val < root.val ? root.left : root.right;
}
return root;
}
}二、二叉搜索树中的插入操作
https://leetcode.cn/problems/insert-into-a-binary-search-tree/?envType=problem-list-v2&envId=8At1GmaZ
class Solution {
public TreeNode insertIntoBST(TreeNode root, int val) {
if(root == null) return new TreeNode(val);
TreeNode node = root;
while(true){
if(node.val < val){
if(node.right == null){
node.right = new TreeNode(val);
break;
}else{
node = node.right;
}
}else{
if(node.left == null){
node.left = new TreeNode(val);
break;
}else{
node = node.left;
}
}
}
return root;
}
}
三、删除二叉搜索树中的节点
https://leetcode.cn/problems/delete-node-in-a-bst/?envType=problem-list-v2&envId=8At1GmaZ
https://img2024.cnblogs.com/blog/2297173/202505/2297173-20250515221923712-465432218.png
class Solution{
if(root == null) return null;
if(key > root.val){
root.right = deleteNode(root.right, key);
}else if(key < root.val){
root.left = deleteNode(root.left, key);
}else{
if(root.left == null) return root.right;
else if(root.right == null)return root.left;
else if(root.left != null && root.right != null){
TreeNode node = root.right; //保存当前节点的右子树
while(node.left != null){
node = node.left; //寻找要删除的节点右子树的最左节点
}
node.left = root.left; // 将欲删除节点的左子树成为其右子树的最左节点的左子树
root = root.right; //欲删除节点的右子顶替其位置，节点被删除
}
}
return root;
}
}
来源：程序园用户自行投稿发布，如果侵权，请联系站长删除
免责声明：如果侵犯了您的权益，请联系站长，我们会及时删除侵权内容，谢谢合作！
页:
[1]