140 lines
8.0 KiB
Markdown
140 lines
8.0 KiB
Markdown
---
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id: 587d8258367417b2b2512c82
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title: Delete a Node with Two Children in a Binary Search Tree
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challengeType: 1
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videoUrl: ''
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localeTitle: 在二叉搜索树中删除具有两个子节点的节点
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---
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## Description
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<section id="description">删除具有两个子节点的节点是最难实现的。删除这样的节点会生成两个不再连接到原始树结构的子树。我们如何重新连接它们？一种方法是在目标节点的右子树中找到最小值，并用该值替换目标节点。以这种方式选择替换确保它大于左子树中的每个节点，它成为新的父节点，但也小于右子树中的每个节点，它成为新的父节点。完成此替换后，必须从右子树中删除替换节点。即使这个操作也很棘手，因为替换可能是一个叶子，或者它本身可能是一个右子树的父亲。如果是叶子，我们必须删除其父对它的引用。否则，它必须是目标的正确子项。在这种情况下，我们必须用替换值替换目标值，并使目标引用替换的右子。说明：让我们通过处理第三种情况来完成我们的<code>remove</code>方法。我们为前两种情况再次提供了一些代码。现在添加一些代码来处理具有两个子节点的目标节点。任何边缘情况要注意？如果树只有三个节点怎么办？完成后，这将完成二进制搜索树的删除操作。干得好，这是一个非常难的问题！ </section>
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## Instructions
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<section id="instructions">
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</section>
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## Tests
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<section id='tests'>
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```yml
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tests:
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- text: 存在<code>BinarySearchTree</code>数据结构。
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testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() }; return (typeof test == "object")})(), "The <code>BinarySearchTree</code> data structure exists.");'
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- text: 二叉搜索树有一个名为<code>remove</code>的方法。
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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>.");'
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- text: 尝试删除不存在的元素将返回<code>null</code> 。
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testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { return false; }; return (typeof test.remove == "function") ? (test.remove(100) == null) : false})(), "Trying to remove an element that does not exist returns <code>null</code>.");'
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- 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> 。
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testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { return false; }; test.add(500); test.remove(500); return (typeof test.remove == "function") ? (test.inorder() == null) : false})(), "If the root node has no children, deleting it sets the root to <code>null</code>.");'
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- text: <code>remove</code>方法从树中删除叶节点
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testString: 'assert((function() { var test = false; if (typeof BinarySearchTree !== "undefined") { test = new BinarySearchTree() } else { 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 (typeof test.remove == "function") ? (test.inorder().join("") == "567") : false})(), "The <code>remove</code> method removes leaf nodes from the tree");'
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- text: <code>remove</code>方法删除具有一个子节点的节点。
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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.");'
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- text: 删除具有两个节点的树中的根将第二个节点设置为根。
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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.");'
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- text: <code>remove</code>方法在保留二叉搜索树结构的同时删除具有两个子节点的节点。
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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(4); test.add(3); test.add(7); test.add(9); test.add(11); test.add(14); test.add(15); test.add(19); test.add(50); test.remove(9); if (!test.isBinarySearchTree()) { return false; }; test.remove(11); if (!test.isBinarySearchTree()) { return false; }; test.remove(14); if (!test.isBinarySearchTree()) { return false; }; test.remove(19); if (!test.isBinarySearchTree()) { return false; }; test.remove(3); if (!test.isBinarySearchTree()) { return false; }; test.remove(50); if (!test.isBinarySearchTree()) { return false; }; test.remove(15); if (!test.isBinarySearchTree()) { return false; }; return (test.inorder().join("") == "147"); })(), "The <code>remove</code> method removes nodes with two children while maintaining the binary search tree structure.");'
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- text: 可以在三个节点的树上删除根。
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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(100); test.add(50); test.add(300); test.remove(100); return (test.inorder().join("") == 50300); })(), "The root can be removed on a tree of three nodes.");'
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```
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</section>
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## Challenge Seed
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<section id='challengeSeed'>
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<div id='js-seed'>
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```js
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var displayTree = (tree) => console.log(JSON.stringify(tree, null, 2));
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function Node(value) {
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this.value = value;
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this.left = null;
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this.right = null;
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}
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function BinarySearchTree() {
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this.root = null;
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this.remove = function(value) {
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if (this.root === null) {
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return null;
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}
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var target;
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var parent = null;
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// find the target value and its parent
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(function findValue(node = this.root) {
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if (value == node.value) {
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target = node;
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} else if (value < node.value && node.left !== null) {
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parent = node;
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return findValue(node.left);
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} else if (value < node.value && node.left === null) {
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return null;
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} else if (value > node.value && node.right !== null) {
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parent = node;
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return findValue(node.right);
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} else {
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return null;
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}
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}).bind(this)();
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if (target === null) {
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return null;
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}
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// count the children of the target to delete
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var children = (target.left !== null ? 1 : 0) + (target.right !== null ? 1 : 0);
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// case 1: target has no children
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if (children === 0) {
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if (target == this.root) {
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this.root = null;
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}
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else {
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if (parent.left == target) {
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parent.left = null;
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} else {
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parent.right = null;
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}
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}
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}
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// case 2: target has one child
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else if (children == 1) {
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var newChild = (target.left !== null) ? target.left : target.right;
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if (parent === null) {
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target.value = newChild.value;
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target.left = null;
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target.right = null;
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} else if (newChild.value < parent.value) {
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parent.left = newChild;
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} else {
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parent.right = newChild;
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}
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target = null;
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}
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// case 3: target has two children, change code below this line
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};
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}
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```
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</div>
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### After Test
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<div id='js-teardown'>
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```js
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console.info('after the test');
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```
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</div>
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</section>
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## Solution
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<section id='solution'>
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```js
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// solution required
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```
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</section>
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