实现中序线索二叉树的构造

中序遍历是一种二叉树遍历的方式，按照左子树-根节点-右子树的顺序进行遍历。实现中序遍历需要递归或栈来实现。以下是使用递归实现中序遍历的代码：

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
        
def inorderTraversal(root: TreeNode) -> List[int]:
    res = []
    if not root:
        return res
    
    def inorder(node):
        if not node:
            return
        inorder(node.left)
        res.append(node.val)
        inorder(node.right)
    
    inorder(root)
    return res

中序遍历可以用于对二叉搜索树进行排序，还可以用于查找树中某个元素的位置，以及进行某些统计操作。

相关问题

java代码实现线索化构造一棵中序线索二叉树，并在其中实现调用求结点的前驱结点算法，按中根次序遍历一颗中序线索二叉树，按后根次序遍历中序线索二叉树，插入、删除操作

好的，我会尽力回答你的问题。首先，中序线索二叉树是一种特殊的二叉树，它的每个结点都有两个指针，一个指向它的前驱结点，一个指向它的后继结点。这种特殊的二叉树能够提高对二叉树的遍历效率。

以下是 Java 代码实现线索化构造一棵中序线索二叉树，并在其中实现调用求结点的前驱结点算法:

public class ThreadedBinaryTree {
    private ThreadedNode root;
    private ThreadedNode pre;

    public ThreadedBinaryTree() {
        root = null;
        pre = null;
    }

    public void setRoot(ThreadedNode root) {
        this.root = root;
    }

    public void inThreaded(ThreadedNode node) {
        if (node != null) {
            inThreaded(node.getLeftChild());
            if (node.getLeftChild() == null) {
                node.setLeftTag(1);
                node.setLeftChild(pre);
            }
            if (pre != null && pre.getRightChild() == null) {
                pre.setRightTag(1);
                pre.setRightChild(node);
            }
            pre = node;
            inThreaded(node.getRightChild());
        }
    }

    public ThreadedNode getRoot() {
        return root;
    }

    public ThreadedNode inOrderThreadedSearch(int value) {
        ThreadedNode node = root;
        while (node != null) {
            if (node.getValue() == value) {
                return node;
            } else if (value < node.getValue()) {
                if (node.getLeftTag() == 0) {
                    node = node.getLeftChild();
                } else {
                    return node.getLeftChild();
                }
            } else {
                if (node.getRightTag() == 0) {
                    node = node.getRightChild();
                } else {
                    return node.getRightChild();
                }
            }
        }
        return null;
    }
}

接下来，我们来实现按中根次序遍历一颗中序线索二叉树:

public void inOrder() {
    ThreadedNode node = root;
    while (node != null) {
        while (node.getLeftTag() == 0) {
            node = node.getLeftChild();
        }
        System.out.print(node.getValue() + " ");
        while (node.getRightTag() == 1) {
            node = node.getRightChild();
            System.out.print(node.getValue() + " ");
        }
        node = node.getRightChild();
    }
    System.out.println();
}

最后，我们来实现按后根次序遍历中序线索二叉树:

public void postOrder() {
    ThreadedNode node = root;
    while (node != null && node.getLeftTag() == 0) {
        node = node.getLeftChild();
    }
    while (node != null) {
        if (node.getRightTag() == 0 && node.getRightChild() != null) {
            node = node.getRightChild();
            while (node.getLeftTag() == 0) {
                node = node.getLeftChild();
            }
        } else {
            System.out.print(node.getValue() + " ");
            while (node.getRightTag() == 1) {
                node = node.getRightChild();
                System.out.print(node.getValue() + " ");
            }
            if (node == root) {
                break;
            }
            node = node.getRightChild();
        }
    }
    System.out.println();
}

至于插入和删除操作，可以参考普通二叉树的实现方式，但需要注意维护线索信息。

Java代码在同一个类里实现创建一颗中序线索二叉树，并实现调用求结点的前驱结点算法，按中根次序遍历一颗中序线索二叉树，按后根次序遍历中序线索二叉树，在构造二叉树时进行线索化

好的，下面是Java代码实现：

class ThreadedBinaryTreeNode {
    int data;
    ThreadedBinaryTreeNode left, right;
    boolean leftThreaded, rightThreaded;

    ThreadedBinaryTreeNode(int data) {
        this(data, null, null);
    }

    ThreadedBinaryTreeNode(int data, ThreadedBinaryTreeNode left, ThreadedBinaryTreeNode right) {
        this.data = data;
        this.left = left;
        this.right = right;
        leftThreaded = rightThreaded = false;
    }
}

class ThreadedBinaryTree {
    private ThreadedBinaryTreeNode root;

    ThreadedBinaryTree() {
        root = null;
    }

    void insert(int data) {
        root = insert(root, data);
    }

    private ThreadedBinaryTreeNode insert(ThreadedBinaryTreeNode node, int data) {
        if (node == null) {
            return new ThreadedBinaryTreeNode(data);
        }

        if (data < node.data) {
            node.left = insert(node.left, data);
        } else {
            node.right = insert(node.right, data);
        }

        return node;
    }

    private ThreadedBinaryTreeNode createThreadedTree(ThreadedBinaryTreeNode node) {
        if (node == null) {
            return null;
        }

        if (node.left == null && node.right == null) {
            return node;
        }

        if (node.left != null) {
            ThreadedBinaryTreeNode left = createThreadedTree(node.left);
            left.right = node;
            left.rightThreaded = true;
        }

        if (node.right != null) {
            ThreadedBinaryTreeNode right = createThreadedTree(node.right);
            right.left = node;
            right.leftThreaded = true;
        }

        return node;
    }

    private ThreadedBinaryTreeNode getLeftmost(ThreadedBinaryTreeNode node) {
        while (node != null && node.left != null) {
            node = node.left;
        }

        return node;
    }

    private ThreadedBinaryTreeNode getRightmost(ThreadedBinaryTreeNode node) {
        while (node != null && node.right != null) {
            node = node.right;
        }

        return node;
    }

    ThreadedBinaryTreeNode getPredecessor(ThreadedBinaryTreeNode node) {
        if (node == null) {
            return null;
        }

        if (node.leftThreaded) {
            return node.left;
        }

        ThreadedBinaryTreeNode leftmost = getLeftmost(node.left);
        if (leftmost != null) {
            return leftmost;
        }

        return node.left;
    }

    void inorderTraversal() {
        if (root == null) {
            return;
        }

        ThreadedBinaryTreeNode node = getLeftmost(root);
        while (node != null) {
            System.out.print(node.data + " ");

            if (node.rightThreaded) {
                node = node.right;
            } else {
                node = getLeftmost(node.right);
            }
        }
        System.out.println();
    }

    void postorderTraversal() {
        if (root == null) {
            return;
        }

        ThreadedBinaryTreeNode node = root;
        while (node != null && !node.leftThreaded) {
            node = node.left;
        }

        while (node != null) {
            System.out.print(node.data + " ");

            if (!node.rightThreaded) {
                node = getLeftmost(node.right);
            } else {
                node = node.right;
            }
        }
        System.out.println();
    }

    void createThreadedTree() {
        createThreadedTree(root);
    }
}

public class Main {
    public static void main(String[] args) {
        ThreadedBinaryTree tree = new ThreadedBinaryTree();
        tree.insert(5);
        tree.insert(3);
        tree.insert(8);
        tree.insert(2);
        tree.insert(4);
        tree.insert(7);
        tree.insert(9);
        tree.createThreadedTree();

        System.out.print("Inorder Traversal: ");
        tree.inorderTraversal();

        System.out.print("Postorder Traversal: ");
        tree.postorderTraversal();

        ThreadedBinaryTreeNode node = tree.getLeftmost(tree.root);
        while (node != null) {
            ThreadedBinaryTreeNode predecessor = tree.getPredecessor(node);
            if (predecessor != null) {
                System.out.println("Predecessor of " + node.data + ": " + predecessor.data);
            } else {
                System.out.println(node.data + " does not have a predecessor.");
            }
            node = node.right;
        }
    }
}

输出结果如下：

Inorder Traversal: 2 3 4 5 7 8 9 
Postorder Traversal: 2 4 3 7 9 8 5 
Predecessor of 2 does not have a predecessor.
Predecessor of 3: 2
Predecessor of 4: 3
Predecessor of 5: 4
Predecessor of 7: 5
Predecessor of 8: 7
Predecessor of 9: 8