235. 二叉搜索树的最近公共祖先
二叉搜索树,顺序查找,找到根节点,
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q); if (root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q); return root; } }
701. 二叉搜索树中的插入操作
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public TreeNode insertIntoBST(TreeNode root, int val) { if (root == null) return new TreeNode(val); TreeNode newRoot = root; TreeNode pre = root; while (root != null) { pre = root; if (root.val > val) { root = root.left; } else if (root.val < val) { root = root.right; } } if (pre.val > val) { pre.left = new TreeNode(val); } else { pre.right = new TreeNode(val); } return newRoot; } }
450. 删除二叉搜索树中的节点
- 第一种情况:没找到删除的节点,遍历到空节点直接返回了
- 找到删除的节点
- 第二种情况:左右孩子都为空(叶子节点),直接删除节点, 返回 NULL 为根节点
- 第三种情况:删除节点的左孩子为空,右孩子不为空,删除节点,右孩子补位,返回右孩子为根节点
- 第四种情况:删除节点的右孩子为空,左孩子不为空,删除节点,左孩子补位,返回左孩子为根节点
- 第五种情况:左右孩子节点都不为空,则将删除节点的左子树头结点(左孩子)放到删除节点的右子树的最左面节点的左孩子上,返回删除节点右孩子为新的根节点。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public TreeNode deleteNode(TreeNode root, int key) { root = delete(root,key); return root; } private TreeNode delete(TreeNode root, int key) { if (root == null) return null; if (root.val > key) { root.left = delete(root.left,key); } else if (root.val < key) { root.right = delete(root.right,key); } else { if (root.left == null) return root.right; if (root.right == null) return root.left; TreeNode tmp = root.right; while (tmp.left != null) { tmp = tmp.left; } root.val = tmp.val; root.right = delete(root.right,tmp.val); } return root; } }