814 Binary Tree Pruning 二叉树剪枝
Description:
Given the root of a binary tree, return the same tree where every subtree (of the given tree) not containing a 1 has been removed.
A subtree of a node node is node plus every node that is a descendant of node.
Example:
Example 1:
Input: root = [1,null,0,0,1]
Output: [1,null,0,null,1]
Explanation:
Only the red nodes satisfy the property "every subtree not containing a 1".
The diagram on the right represents the answer.
Example 2:
Input: root = [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]
Example 3:
Input: root = [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]
Constraints:
The number of nodes in the tree is in the range [1, 200].
Node.val is either 0 or 1.
题目描述:
给定二叉树根结点 root ，此外树的每个结点的值要么是 0，要么是 1。
返回移除了所有不包含 1 的子树的原二叉树。
( 节点 X 的子树为 X 本身，以及所有 X 的后代。)
示例 :
示例1:
输入: [1,null,0,0,1]
输出: [1,null,0,null,1]
解释:
只有红色节点满足条件“所有不包含 1 的子树”。
右图为返回的答案。
示例2:
输入: [1,0,1,0,0,0,1]
输出: [1,null,1,null,1]
示例3:
输入: [1,1,0,1,1,0,1,0]
输出: [1,1,0,1,1,null,1]
说明:
给定的二叉树最多有 200 个节点。
每个节点的值只会为 0 或 1 。
思路:
后序遍历
要删除 root 首先要删除 root -> left, root -> right
先判断 root 是否为空指针, 是的话直接返回自身
然后判断 root 是否为叶结点且 root -> val == 0, 直接进行删除操作即可
时间复杂度为 O(n), 空间复杂度为 O(h), 其中 h 为树的高度
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* pruneTree(TreeNode* root)
{
if (!root) return root;
root -> left = pruneTree(root -> left);
root -> right = pruneTree(root -> right);
return !root -> left and !root -> right and !root -> val ? 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 pruneTree(TreeNode root) {
if (root == null) return root;
root.left = pruneTree(root.left);
root.right = pruneTree(root.right);
return root.left == null && root.right == null && root.val == 0 ? 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 pruneTree(self, root: TreeNode) -> TreeNode:
if not root:
return root
root.left, root.right = self.pruneTree(root.left), self.pruneTree(root.right)
return None if not root.left and not root.right and not root.val else root
