freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/data-structures/delete-a-node-with-two-children-in-a-binary-search-tree.md

---
id: 587d8258367417b2b2512c82
title: 在二叉搜索树中删除具有两个子节点的节点
challengeType: 1
videoUrl: ''
dashedName: delete-a-node-with-two-children-in-a-binary-search-tree
---

# --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`remove`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！

# --hints--

存在`BinarySearchTree`数据结构。

```js
assert(
  (function () {
    var test = false;
    if (typeof BinarySearchTree !== 'undefined') {
      test = new BinarySearchTree();
    }
    return typeof test == 'object';
  })()
);
```

二叉搜索树有一个名为`remove`的方法。

```js
assert(
  (function () {
    var test = false;
    if (typeof BinarySearchTree !== 'undefined') {
      test = new BinarySearchTree();
    } else {
      return false;
    }
    return typeof test.remove == 'function';
  })()
);
```

尝试删除不存在的元素将返回`null` 。

```js
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;
  })()
);
```

如果根节点没有子节点，则删除它会将根节点设置为`null` 。

```js
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;
  })()
);
```

`remove`方法从树中删除叶节点

```js
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;
  })()
);
```

`remove`方法删除具有一个子节点的节点。

```js
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';
  })()
);
```

删除具有两个节点的树中的根将第二个节点设置为根。

```js
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';
  })()
);
```

`remove`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。

```js
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';
  })()
);
```

可以在三个节点的树上删除根。

```js
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;
  })()
);
```

# --seed--

## --after-user-code--

```js
BinarySearchTree.prototype = Object.assign(
  BinarySearchTree.prototype,
  {
    add: function(value) {
      var node = this.root;
      if (node == null) {
        this.root = new Node(value);
        return;
      } else {
        function searchTree(node) {
          if (value < node.value) {
            if (node.left == null) {
              node.left = new Node(value);
              return;
            } else if (node.left != null) {
              return searchTree(node.left);
            }
          } else if (value > node.value) {
            if (node.right == null) {
              node.right = new Node(value);
              return;
            } else if (node.right != null) {
              return searchTree(node.right);
            }
          } else {
            return null;
          }
        }
        return searchTree(node);
      }
    },
    inorder: function() {
      if (this.root == null) {
        return null;
      } else {
        var result = new Array();
        function traverseInOrder(node) {
          if (node.left != null) {
            traverseInOrder(node.left);
          }
          result.push(node.value);
          if (node.right != null) {
            traverseInOrder(node.right);
          }
        }
        traverseInOrder(this.root);
        return result;
      }
    },
    isBinarySearchTree() {
      if (this.root == null) {
        return null;
      } else {
        var check = true;
        function checkTree(node) {
          if (node.left != null) {
            var left = node.left;
            if (left.value > node.value) {
              check = false;
            } else {
              checkTree(left);
            }
          }
          if (node.right != null) {
            var right = node.right;
            if (right.value < node.value) {
              check = false;
            } else {
              checkTree(right);
            }
          }
        }
        checkTree(this.root);
        return check;
      }
    }
  }
);
```

## --seed-contents--

```js
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
    else if (children == 1) {
      var newChild = target.left !== null ? target.left : target.right;
      if (parent === null) {
        target.value = newChild.value;
        target.left = null;
        target.right = null;
      } else if (newChild.value < parent.value) {
        parent.left = newChild;
      } else {
        parent.right = newChild;
      }
      target = null;
    }
    // Case 3: Target has two children
    // Only change code below this line
  };
}
```

# --solutions--

```js
// solution required
```