Verify Preorder Serialization of a Binary Tree

本文介绍了一种方法来判断给定的字符串是否为一棵二叉树的有效前序遍历序列。通过两种算法——出度&入度及栈操作实现验证,并详细解析了每种方法的工作原理。

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题目:

给定一棵二叉树的前序遍历序列,如 "9,3,4,#,#,1,#,#,2,#,6,#,#"  ,# 代表空节点,判断是否是一个正确的二叉树前序遍历序列


提供两种解题思路,出度&入度, 栈


1. 出度& 入度

/*
    这道题使用出度和入度来解决,对于以下节点来说:
    非空节点: 出度 2, 入度 1
    空节点:   出度 0, 入度 1
    那么当我们建立一棵二叉树的时候,出度和入度之差 diff = 出度 - 入度
    对于加入任何一个节点,diff - 1, 因为整体的树为它减少了一个入度。
    当加入一个非空节点的时候,diff + 2, 因为它提供了两个出度。
    那么当树是正确的时候,diff的值应该为非负。如果为正确的树的时候,
    diff == 0
    */
    public boolean isValidSerialization(String preorder) {
        String[] nodes = preorder.split(",");
        int diff = 1;
        for (int i = 0; i < nodes.length; ++i) {
            //先减入度,再加出度
            if (--diff < 0) {
                return false;
            }
            if (!"#".equals(nodes[i])) {
                diff += 2;
            }
        }
        return diff == 0;
    }

2. 栈

/*
    已例子一为例,:”9,3,4,#,#,1,#,#,2,#,6,#,#” 遇到x # #的时候,就把它变为 #
    模拟一遍过程:

    9,3,4,#,# => 9,3,# 继续读
    9,3,#,1,#,# => 9,3,#,# => 9,# 继续读
    9,#2,#,6,#,# => 9,#,2,#,# => 9,#,# => #
    栈中最后一个元素为 # 的时候结果一定是true
    */
    public boolean isValidSerialization(String preorder) {
        
        String[] nodes = preorder.split(",");
        ArrayList<String> stack = new ArrayList<String>();

        for (int i = 0; i < nodes.length; ++i) {
            
            stack.add(nodes[i]);
            
            while (stack.size() >= 3 
                && stack.get(stack.size() - 1 ).equals("#")
                && stack.get(stack.size() - 2 ).equals("#")
                && !stack.get(stack.size() - 3 ).equals("#")) {
                
                stack.remove(stack.size() - 1);
                stack.remove(stack.size() - 1);
                stack.remove(stack.size() - 1);
                
                stack.add("#");
            }
        }
     
        if( stack.size() == 1 && stack.get(0).equals("#"))
            return true;
        else
            return false;
    }


--------------EOF---------------

【Solution】 To convert a binary search tree into a sorted circular doubly linked list, we can use the following steps: 1. Inorder traversal of the binary search tree to get the elements in sorted order. 2. Create a doubly linked list and add the elements from the inorder traversal to it. 3. Make the list circular by connecting the head and tail nodes. 4. Return the head node of the circular doubly linked list. Here's the Python code for the solution: ``` class Node: def __init__(self, val): self.val = val self.prev = None self.next = None def tree_to_doubly_list(root): if not root: return None stack = [] cur = root head = None prev = None while cur or stack: while cur: stack.append(cur) cur = cur.left cur = stack.pop() if not head: head = cur if prev: prev.right = cur cur.left = prev prev = cur cur = cur.right head.left = prev prev.right = head return head ``` To verify the accuracy of the code, we can use the following test cases: ``` # Test case 1 # Input: [4,2,5,1,3] # Output: # Binary search tree: # 4 # / \ # 2 5 # / \ # 1 3 # Doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> 5 # Doubly linked list in reverse order: 5 <-> 4 <-> 3 <-> 2 <-> 1 root = Node(4) root.left = Node(2) root.right = Node(5) root.left.left = Node(1) root.left.right = Node(3) head = tree_to_doubly_list(root) print("Binary search tree:") print_tree(root) print("Doubly linked list:") print_list(head) print("Doubly linked list in reverse order:") print_list_reverse(head) # Test case 2 # Input: [2,1,3] # Output: # Binary search tree: # 2 # / \ # 1 3 # Doubly linked list: 1 <-> 2 <-> 3 # Doubly linked list in reverse order: 3 <-> 2 <-> 1 root = Node(2) root.left = Node(1) root.right = Node(3) head = tree_to_doubly_list(root) print("Binary search tree:") print_tree(root) print("Doubly linked list:") print_list(head) print("Doubly linked list in reverse order:") print_list_reverse(head) ``` The output of the test cases should match the expected output as commented in the code.
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