1325 Delete Leaves With a Given Value 删除给定值的叶子节点
Description:
Given a binary tree root and an integer target, delete all the leaf nodes with value target.
Note that once you delete a leaf node with value target, if its parent node becomes a leaf node and has the value target, it should also be deleted (you need to continue doing that until you cannot).
Example:
Example 1:
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Input: root = [1,2,3,2,null,2,4], target = 2
Output: [1,null,3,null,4]
Explanation: Leaf nodes in green with value (target = 2) are removed (Picture in left).
After removing, new nodes become leaf nodes with value (target = 2) (Picture in center).
Example 2:
[图片上传失败...(image-c90db3-1664117570123)]
Input: root = [1,3,3,3,2], target = 3
Output: [1,3,null,null,2]
Example 3:
[图片上传失败...(image-a992ce-1664117570123)]
Input: root = [1,2,null,2,null,2], target = 2
Output: [1]
Explanation: Leaf nodes in green with value (target = 2) are removed at each step.
Constraints:
The number of nodes in the tree is in the range [1, 3000].
1 <= Node.val, target <= 1000
题目描述：
给你一棵以 root 为根的二叉树和一个整数 target ，请你删除所有值为 target 的 叶子节点 。
注意，一旦删除值为 target 的叶子节点，它的父节点就可能变成叶子节点；如果新叶子节点的值恰好也是 target ，那么这个节点也应该被删除。
也就是说，你需要重复此过程直到不能继续删除。
示例：
示例 1：
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输入：root = [1,2,3,2,null,2,4], target = 2
输出：[1,null,3,null,4]
解释：
上面左边的图中，绿色节点为叶子节点，且它们的值与 target 相同（同为 2 ），它们会被删除，得到中间的图。
有一个新的节点变成了叶子节点且它的值与 target 相同，所以将再次进行删除，从而得到最右边的图。
示例 2：
[图片上传失败...(image-9d7d13-1664117570123)]
输入：root = [1,3,3,3,2], target = 3
输出：[1,3,null,null,2]
示例 3：
[图片上传失败...(image-853d45-1664117570123)]
输入：root = [1,2,null,2,null,2], target = 2
输出：[1]
解释：每一步都删除一个绿色的叶子节点（值为 2）。
示例 4：
输入：root = [1,1,1], target = 1
输出：[]
示例 5：
输入：root = [1,2,3], target = 1
输出：[1,2,3]
提示：
1 <= target <= 1000
每一棵树最多有 3000 个节点。
每一个节点值的范围是 [1, 1000] 。
思路：
递归
当前结点为空返回空
否则后序遍历二叉树
找到所有叶子结点的值为 target 的置空
时间复杂度为 O(n), 空间复杂度为 O(n)
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution
{
public:
TreeNode* removeLeafNodes(TreeNode* root, int target)
{
if (!root) return nullptr;
root -> left = removeLeafNodes(root -> left, target);
root -> right = removeLeafNodes(root -> right, target);
return (root -> val == target and !root -> left and !root -> right) ? nullptr : root;
}
};
Java:
/**
* 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 removeLeafNodes(TreeNode root, int target) {
if (root == null) return null;
root.left = removeLeafNodes(root.left, target);
root.right = removeLeafNodes(root.right, target);
return (root.val == target && root.left == null && root.right == null) ? null : root;
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def removeLeafNodes(self, root: Optional[TreeNode], target: int) -> Optional[TreeNode]:
if not root:
return None
root.left, root.right = self.removeLeafNodes(root.left, target), self.removeLeafNodes(root.right, target)
return None if root.val == target and not root.left and not root.right else root
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