剑指 Offer II 047. 二叉树剪枝

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剑指 Offer II 047. 二叉树剪枝

题目描述

给定一个二叉树 根节点 root ，树的每个节点的值要么是 0，要么是 1。请剪除该二叉树中所有节点的值为 0 的子树。

节点 node 的子树为 node 本身，以及所有 node 的后代。

 

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]
解释:

示例 3:

输入: [1,1,0,1,1,0,1,0]
输出: [1,1,0,1,1,null,1]
解释:

 

提示:

• 二叉树的节点个数的范围是 [1,200]
• 二叉树节点的值只会是 0 或 1

 

注意：本题与主站 814 题相同：https://leetcode.cn/problems/binary-tree-pruning/

解法

方法一

Python3

# 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 None root.left=self.pruneTree(root.left) #左子树净化root.right=self.pruneTree(root.right) #右子树净化if root.val==0 and not root.left and not root.right:return None #根子树净化return 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 null;}root.left = pruneTree(root.left);root.right = pruneTree(root.right);if (root.val == 0 && root.left == null && root.right == null) {return null;}return root;}
}

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 nullptr;root->left = pruneTree(root->left);root->right = pruneTree(root->right);if (!root->val && !root->left && !root->right) return nullptr;return root;}
};

Go

/*** Definition for a binary tree node.* type TreeNode struct {*     Val int*     Left *TreeNode*     Right *TreeNode* }*/
func pruneTree(root *TreeNode) *TreeNode {if root == nil {return nil}root.Left = pruneTree(root.Left)root.Right = pruneTree(root.Right)if root.Val == 0 && root.Left == nil && root.Right == nil {return nil}return root
}

JavaScript

/*** Definition for a binary tree node.* function TreeNode(val, left, right) {*     this.val = (val===undefined ? 0 : val)*     this.left = (left===undefined ? null : left)*     this.right = (right===undefined ? null : right)* }*/
/*** @param {TreeNode} root* @return {TreeNode}*/
var pruneTree = function (root) {if (!root) return null;root.left = pruneTree(root.left);root.right = pruneTree(root.right);if (root.val == 0 && !root.left && !root.right) {return null;}return root;
};

Swift

/* class TreeNode {
*     var val: Int
*     var left: TreeNode?
*     var right: TreeNode?
*     init() {
*         self.val = 0
*         self.left = nil
*         self.right = nil
*     }
*     init(_ val: Int) {
*         self.val = val
*         self.left = nil
*         self.right = nil
*     }
*     init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
*         self.val = val
*         self.left = left
*         self.right = right
*     }
* }
*/class Solution {func pruneTree(_ root: TreeNode?) -> TreeNode? {guard let root = root else {return nil}root.left = pruneTree(root.left)root.right = pruneTree(root.right)if root.val == 0 && root.left == nil && root.right == nil {return nil}return root}
}