226. Invert Binary Tree

LeetCode easy original: C# #backtracking #csharp #easy #leetcode #recursion #tree #two-pointers

O texto da tarefa é traduzido do russo para o idioma selecionado. O código permanece sem alterações.

LeetCode Static

Дан корень бинарного дерева, инвертируйте árvore и return его корень.

Exemplo:

Input: root = [4,2,7,1,3,6,9]

Output: [4,7,2,9,6,3,1]

C# solução

correspondente/original

public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int x) { val = x; }
}
public class Solution {
    public TreeNode InvertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        TreeNode right = InvertTree(root.right);
        TreeNode left = InvertTree(root.left);
        root.left = right;
        root.right = left;
        return root;
    }
}

C++ solução

rascunho automático, revisar antes de enviar

#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int x) { val = x; }
}
class Solution {
public:
    public TreeNode InvertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        TreeNode right = InvertTree(root.right);
        TreeNode left = InvertTree(root.left);
        root.left = right;
        root.right = left;
        return root;
    }
}

Java solução

correspondente/original

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        TreeNode right = invertTree(root.right);
        TreeNode left = invertTree(root.left);
        root.left = right;
        root.right = left;
        return root;
    }
}

JavaScript solução

correspondente/original

function TreeNode(val, left, right) {
    this.val = (val === undefined ? 0 : val)
    this.left = (left === undefined ? null : left)
    this.right = (right === undefined ? null : right)
}

var invertTree = function(root) {
    if (root === null) {
        return null;
    }
    const right = invertTree(root.right);
    const left = invertTree(root.left);
    root.left = right;
    root.right = left;
    return root;
};

Python solução

correspondente/original

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        if root is None:
            return None
        right = self.invertTree(root.right)
        left = self.invertTree(root.left)
        root.left = right
        root.right = left
        return root

Go solução

correspondente/original

package main

import "fmt"

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func invertTree(root *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    right := invertTree(root.Right)
    left := invertTree(root.Left)
    root.Left = right
    root.Right = left
    return root
}

Algorithm

Рекурсивная инверсия поддеревьев:

Если текущий узел (root) пустой, возвращаем null.

Рекурсивно инвертируем правое подárvore и сохраняем результат в переменную right.

Рекурсивно инвертируем левое подárvore и сохраняем результат в переменную left.

Обмен поддеревьев:

Устанавливаем левое подárvore текущего узла равным правому поддереву.

Устанавливаем правое подárvore текущего узла равным левому поддереву.

Возврат корня:

Возвращаем текущий узел (root) как новый корень инвертированного дерева.

😎

Vacancies for this task

vagas ativas with overlapping task tags are mostradas.

Todas as vagas

Ainda não há vagas ativas.