5.9 KiB
5.9 KiB
id | title | challengeType | videoUrl | localeTitle |
---|---|---|---|---|
587d8258367417b2b2512c81 | Delete a Node with One Child in a Binary Search Tree | 1 | 在二叉搜索树中删除具有一个子节点的节点 |
Description
remove
方法中提供了一些代码,用于完成上一次挑战中的任务。我们找到要删除的目标及其父节点,并定义目标节点具有的子节点数。让我们在这里为仅有一个子节点的目标节点添加下一个案例。在这里,我们必须确定单个子节点是树中的左或右分支,然后在父节点中设置正确的引用以指向此节点。另外,让我们考虑目标是根节点的情况(这意味着父节点将为null
)。只要通过测试,请随意用自己的代码替换所有入门代码。 Instructions
Tests
tests:
- text: 存在<code>BinarySearchTree</code>数据结构。
testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() }; return (typeof test == "object")})(), "The <code>BinarySearchTree</code> data structure exists.");'
- text: 二叉搜索树有一个名为<code>remove</code>的方法。
testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { return false; }; return (typeof test.remove == "function")})(), "The binary search tree has a method called <code>remove</code>.");'
- text: 尝试删除不存在的元素将返回<code>null</code> 。
testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { return false; }; if (typeof test.remove !== "function") { return false; }; return (test.remove(100) == null); })(), "Trying to remove an element that does not exist returns <code>null</code>.");'
- text: 如果根节点没有子节点,则删除它会将根节点设置为<code>null</code> 。
testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { return false; }; if (typeof test.remove !== "function") { return false; }; test.add(500); test.remove(500); return (test.inorder() == null); })(), "If the root node has no children, deleting it sets the root to <code>null</code>.");'
- text: <code>remove</code>方法从树中删除叶节点
testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { return false; }; if (typeof test.remove !== "function") { return false; }; test.add(5); test.add(3); test.add(7); test.add(6); test.add(10); test.add(12); test.remove(3); test.remove(12); test.remove(10); return (test.inorder().join("") == "567"); })(), "The <code>remove</code> method removes leaf nodes from the tree");'
- text: <code>remove</code>方法删除具有一个子节点的节点。
testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { return false; }; if (typeof test.remove !== "function") { return false; }; test.add(-1); test.add(3); test.add(7); test.add(16); test.remove(16); test.remove(7); test.remove(3); return (test.inorder().join("") == "-1"); })(), "The <code>remove</code> method removes nodes with one child.");'
- text: 删除具有两个节点的树中的根将第二个节点设置为根。
testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { return false; }; if (typeof test.remove !== "function") { return false; }; test.add(15); test.add(27); test.remove(15); return (test.inorder().join("") == "27"); })(), "Removing the root in a tree with two nodes sets the second to be the root.");'
Challenge Seed
var displayTree = (tree) => console.log(JSON.stringify(tree, null, 2));
function Node(value) {
this.value = value;
this.left = null;
this.right = null;
}
function BinarySearchTree() {
this.root = null;
this.remove = function(value) {
if (this.root === null) {
return null;
}
var target;
var parent = null;
// find the target value and its parent
(function findValue(node = this.root) {
if (value == node.value) {
target = node;
} else if (value < node.value && node.left !== null) {
parent = node;
return findValue(node.left);
} else if (value < node.value && node.left === null) {
return null;
} else if (value > node.value && node.right !== null) {
parent = node;
return findValue(node.right);
} else {
return null;
}
}).bind(this)();
if (target === null) {
return null;
}
// count the children of the target to delete
var children = (target.left !== null ? 1 : 0) + (target.right !== null ? 1 : 0);
// case 1: target has no children
if (children === 0) {
if (target == this.root) {
this.root = null;
}
else {
if (parent.left == target) {
parent.left = null;
} else {
parent.right = null;
}
}
}
// case 2: target has one child, change code below this line
};
}
After Test
console.info('after the test');
Solution
// solution required