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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
}
}
