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中等
数组
分治
矩阵

English Version

题目描述

给你一个 n * n 矩阵 grid ,矩阵由若干 01 组成。请你用四叉树表示该矩阵 grid

你需要返回能表示矩阵 grid 的 四叉树 的根结点。

四叉树数据结构中,每个内部节点只有四个子节点。此外,每个节点都有两个属性:

  • val:储存叶子结点所代表的区域的值。1 对应 True,0 对应 False。注意,当 isLeafFalse 时,你可以把 True 或者 False 赋值给节点,两种值都会被判题机制 接受
  • isLeaf: 当这个节点是一个叶子结点时为 True,如果它有 4 个子节点则为 False
class Node {
    public boolean val;
    public boolean isLeaf;
    public Node topLeft;
    public Node topRight;
    public Node bottomLeft;
    public Node bottomRight;
}

我们可以按以下步骤为二维区域构建四叉树:

  1. 如果当前网格的值相同(即,全为 0 或者全为 1),将 isLeaf 设为 True ,将 val 设为网格相应的值,并将四个子节点都设为 Null 然后停止。
  2. 如果当前网格的值不同,将 isLeaf 设为 False, 将 val 设为任意值,然后如下图所示,将当前网格划分为四个子网格。
  3. 使用适当的子网格递归每个子节点。

如果你想了解更多关于四叉树的内容,可以参考 wiki

四叉树格式:

你不需要阅读本节来解决这个问题。只有当你想了解输出格式时才会这样做。输出为使用层序遍历后四叉树的序列化形式,其中 null 表示路径终止符,其下面不存在节点。

它与二叉树的序列化非常相似。唯一的区别是节点以列表形式表示 [isLeaf, val]

如果 isLeaf 或者 val 的值为 True ,则表示它在列表 [isLeaf, val] 中的值为 1 ;如果 isLeaf 或者 val 的值为 False ,则表示值为 0

 

示例 1:

输入:grid = [[0,1],[1,0]]
输出:[[0,1],[1,0],[1,1],[1,1],[1,0]]
解释:此示例的解释如下:
请注意,在下面四叉树的图示中,0 表示 false,1 表示 True 。

示例 2:

输入:grid = [[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,1,1,1,1],[1,1,1,1,1,1,1,1],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0],[1,1,1,1,0,0,0,0]]
输出:[[0,1],[1,1],[0,1],[1,1],[1,0],null,null,null,null,[1,0],[1,0],[1,1],[1,1]]
解释:网格中的所有值都不相同。我们将网格划分为四个子网格。
topLeft,bottomLeft 和 bottomRight 均具有相同的值。
topRight 具有不同的值,因此我们将其再分为 4 个子网格,这样每个子网格都具有相同的值。
解释如下图所示:

 

提示:

  1. n == grid.length == grid[i].length
  2. n == 2x 其中 0 <= x <= 6

解法

方法一:DFS

DFS 递归遍历 grid,先判断 grid 是否为叶子节点,是则返回叶子节点相关信息;否则递归 grid 4 个子节点。

Python3

"""
# Definition for a QuadTree node.
class Node:
    def __init__(self, val, isLeaf, topLeft, topRight, bottomLeft, bottomRight):
        self.val = val
        self.isLeaf = isLeaf
        self.topLeft = topLeft
        self.topRight = topRight
        self.bottomLeft = bottomLeft
        self.bottomRight = bottomRight
"""


class Solution:
    def construct(self, grid: List[List[int]]) -> 'Node':
        def dfs(a, b, c, d):
            zero = one = 0
            for i in range(a, c + 1):
                for j in range(b, d + 1):
                    if grid[i][j] == 0:
                        zero = 1
                    else:
                        one = 1
            isLeaf = zero + one == 1
            val = isLeaf and one
            if isLeaf:
                return Node(grid[a][b], True)
            topLeft = dfs(a, b, (a + c) // 2, (b + d) // 2)
            topRight = dfs(a, (b + d) // 2 + 1, (a + c) // 2, d)
            bottomLeft = dfs((a + c) // 2 + 1, b, c, (b + d) // 2)
            bottomRight = dfs((a + c) // 2 + 1, (b + d) // 2 + 1, c, d)
            return Node(val, isLeaf, topLeft, topRight, bottomLeft, bottomRight)

        return dfs(0, 0, len(grid) - 1, len(grid[0]) - 1)

Java

/*
// Definition for a QuadTree node.
class Node {
    public boolean val;
    public boolean isLeaf;
    public Node topLeft;
    public Node topRight;
    public Node bottomLeft;
    public Node bottomRight;


    public Node() {
        this.val = false;
        this.isLeaf = false;
        this.topLeft = null;
        this.topRight = null;
        this.bottomLeft = null;
        this.bottomRight = null;
    }

    public Node(boolean val, boolean isLeaf) {
        this.val = val;
        this.isLeaf = isLeaf;
        this.topLeft = null;
        this.topRight = null;
        this.bottomLeft = null;
        this.bottomRight = null;
    }

    public Node(boolean val, boolean isLeaf, Node topLeft, Node topRight, Node bottomLeft, Node
bottomRight) { this.val = val; this.isLeaf = isLeaf; this.topLeft = topLeft; this.topRight =
topRight; this.bottomLeft = bottomLeft; this.bottomRight = bottomRight;
    }
};
*/

class Solution {
    public Node construct(int[][] grid) {
        return dfs(0, 0, grid.length - 1, grid[0].length - 1, grid);
    }

    private Node dfs(int a, int b, int c, int d, int[][] grid) {
        int zero = 0, one = 0;
        for (int i = a; i <= c; ++i) {
            for (int j = b; j <= d; ++j) {
                if (grid[i][j] == 0) {
                    zero = 1;
                } else {
                    one = 1;
                }
            }
        }
        boolean isLeaf = zero + one == 1;
        boolean val = isLeaf && one == 1;
        Node node = new Node(val, isLeaf);
        if (isLeaf) {
            return node;
        }
        node.topLeft = dfs(a, b, (a + c) / 2, (b + d) / 2, grid);
        node.topRight = dfs(a, (b + d) / 2 + 1, (a + c) / 2, d, grid);
        node.bottomLeft = dfs((a + c) / 2 + 1, b, c, (b + d) / 2, grid);
        node.bottomRight = dfs((a + c) / 2 + 1, (b + d) / 2 + 1, c, d, grid);
        return node;
    }
}

C++

/*
// Definition for a QuadTree node.
class Node {
public:
    bool val;
    bool isLeaf;
    Node* topLeft;
    Node* topRight;
    Node* bottomLeft;
    Node* bottomRight;

    Node() {
        val = false;
        isLeaf = false;
        topLeft = NULL;
        topRight = NULL;
        bottomLeft = NULL;
        bottomRight = NULL;
    }

    Node(bool _val, bool _isLeaf) {
        val = _val;
        isLeaf = _isLeaf;
        topLeft = NULL;
        topRight = NULL;
        bottomLeft = NULL;
        bottomRight = NULL;
    }

    Node(bool _val, bool _isLeaf, Node* _topLeft, Node* _topRight, Node* _bottomLeft, Node* _bottomRight) {
        val = _val;
        isLeaf = _isLeaf;
        topLeft = _topLeft;
        topRight = _topRight;
        bottomLeft = _bottomLeft;
        bottomRight = _bottomRight;
    }
};
*/

class Solution {
public:
    Node* construct(vector<vector<int>>& grid) {
        return dfs(0, 0, grid.size() - 1, grid[0].size() - 1, grid);
    }

    Node* dfs(int a, int b, int c, int d, vector<vector<int>>& grid) {
        int zero = 0, one = 0;
        for (int i = a; i <= c; ++i) {
            for (int j = b; j <= d; ++j) {
                if (grid[i][j])
                    one = 1;
                else
                    zero = 1;
            }
        }
        bool isLeaf = zero + one == 1;
        bool val = isLeaf && one;
        Node* node = new Node(val, isLeaf);
        if (isLeaf) return node;
        node->topLeft = dfs(a, b, (a + c) / 2, (b + d) / 2, grid);
        node->topRight = dfs(a, (b + d) / 2 + 1, (a + c) / 2, d, grid);
        node->bottomLeft = dfs((a + c) / 2 + 1, b, c, (b + d) / 2, grid);
        node->bottomRight = dfs((a + c) / 2 + 1, (b + d) / 2 + 1, c, d, grid);
        return node;
    }
};

Go

/**
 * Definition for a QuadTree node.
 * type Node struct {
 *     Val bool
 *     IsLeaf bool
 *     TopLeft *Node
 *     TopRight *Node
 *     BottomLeft *Node
 *     BottomRight *Node
 * }
 */

func construct(grid [][]int) *Node {
	var dfs func(a, b, c, d int) *Node
	dfs = func(a, b, c, d int) *Node {
		zero, one := 0, 0
		for i := a; i <= c; i++ {
			for j := b; j <= d; j++ {
				if grid[i][j] == 0 {
					zero = 1
				} else {
					one = 1
				}
			}
		}
		isLeaf := zero+one == 1
		val := isLeaf && one == 1
		node := &Node{Val: val, IsLeaf: isLeaf}
		if isLeaf {
			return node
		}
		node.TopLeft = dfs(a, b, (a+c)/2, (b+d)/2)
		node.TopRight = dfs(a, (b+d)/2+1, (a+c)/2, d)
		node.BottomLeft = dfs((a+c)/2+1, b, c, (b+d)/2)
		node.BottomRight = dfs((a+c)/2+1, (b+d)/2+1, c, d)
		return node
	}
	return dfs(0, 0, len(grid)-1, len(grid[0])-1)
}