javascript - 如何更改二叉树索引中递归的保留?
问题描述
我用 JavaScript 编写了允许生成任何二叉树的代码(每个父母都有 2 个孩子)。
用户可以通过称为“级别”的变量指定他想要的树的多少级别。
代码:
function TreeNode(name, type, properties, children) {
this.name = name;
this.type = type;
this.properties = properties;
this.children = children;
}
var levels = 3;
prop = properties("Object_1");
var tree = new TreeNode("Object_1", "level_1", prop, []);
levels--;
var tempIndex = 2;
function generateArbitraryLevels(parent, levelsRemaining) {
// last level
if (levelsRemaining === 0) return;
var currentLevel = parseInt(parent.type.split('_')[1]) + 1;
var prop1 = properties("Object_" + tempIndex);
parent.children.push(
new TreeNode("Object_" + tempIndex++, "level_" + currentLevel, prop1, [])
);
generateArbitraryLevels(parent.children[0], levelsRemaining - 1);
var prop2 = properties("Object_" + tempIndex);
parent.children.push(
new TreeNode("Object_" + tempIndex++, "level_" + currentLevel, prop2, [])
);
generateArbitraryLevels(parent.children[1], levelsRemaining - 1);
}
generateArbitraryLevels(tree, levels);
tree = JSON.stringify([tree]);
树中的每个对象都有:
-name - 架构中的第一个字段,它应该包括对象的 ID,
-type - 第二个字段,它包含有关对象级别的信息,
- 属性- 没关系,
-儿童。
所以索引应该从左到右。我怎样才能做到这一点?也许循环会更好地达到这个目的?
解决方案
我相信你正在寻找Breadth First
而不是Depth First
你已经实施。检查下面的链接以获取有关树中广度优先算法的更多信息https://www.cs.bu.edu/teaching/c/tree/breadth-first/
您可以参考以下代码以获取用例的示例实现
const TreeNode = (name, type, properties, children) => ({
name,
type,
properties,
children,
});
const getNode = (index, level, properties) => {
const name = `Object_${index}`;
const type = `level_${level}`;
return TreeNode(name, type, properties[name], []);
};
const generateTreeChildren = (node, currentLevel, maxLevels, index, properties) => {
if (currentLevel > maxLevels) {
return null;
}
const child1 = getNode(index, currentLevel, properties);
const child2 = getNode(index + 1, currentLevel, properties);
node.children = [child1, child2];
generateTreeChildren(child1, currentLevel + 1, maxLevels, index + 2, properties);
generateTreeChildren(child2, currentLevel + 1, maxLevels, index + 4, properties);
};
const generateTree = (maxLevels, properties) => {
if (maxLevels === 0) {
return null;
}
const node = getNode(1, 1, properties);
generateTreeChildren(node, 2, maxLevels, 2, properties);
return node;
};
console.info(JSON.stringify(generateTree(3, {})));