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上次编辑到这里,代码来自缓存 点击恢复默认模板
/**
* 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) {
}
};
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golang 解法, 执行用时: 4 ms, 内存消耗: 5.2 MB, 提交时间: 2022-11-25 21:21:23
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func removeLeafNodes(root *TreeNode, target int) *TreeNode {
if root == nil {
return root
}
root.Left, root.Right = removeLeafNodes(root.Left, target), removeLeafNodes(root.Right, target)
if root.Left == nil && root.Right == nil && root.Val == target {
return nil
}
return root
}
python3 解法, 执行用时: 68 ms, 内存消耗: 15.6 MB, 提交时间: 2022-11-25 21:19:40
# 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: TreeNode, target: int) -> TreeNode:
if not root:
return None
root.left = self.removeLeafNodes(root.left, target)
root.right = self.removeLeafNodes(root.right, target)
if not root.left and not root.right and root.val == target:
return None
return root