331. Verify Preorder Serialization of a Binary Tree(难)

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as #.

     _9_
    /   \
   3     2
  / \   / \
 4   1  #  6
/ \ / \   / \
# # # #   # #

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character '#' representing null pointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".

Example 1:
"9,3,4,#,#,1,#,#,2,#,6,#,#"
Return true

Example 2:
"1,#"
Return false

Example 3:
"9,#,#,1"

Return false

一、

将元素依次push到vector,如果vector的大小大于等于3,且后两个为#,倒数第三个不为#,则弹出他们,push一个#。

如果最后vector大小为1,且为#,则是合法的前序序列!

class Solution {
public:
	bool isValidSerialization(string preorder) {
		vector<string> vec;
		stringstream ss(preorder);
		string val;
		while (getline(ss, val, ',')){
			vec.push_back(val);
			while (vec.size() >= 3 && vec[vec.size() - 1] == "#"
				&&vec[vec.size() - 2] == "#" && vec[vec.size() - 3] != "#"){
				vec.pop_back(); vec.pop_back(); vec.pop_back();
				vec.push_back("#");
			}
		}
		return vec.size() == 1 && vec[0] == "#";
	}
};

二、

在构建的过程中,记录出度与入度之差,记为diff = outdegree - indegree

当遍历节点时,我们令diff - 1(因为节点提供了一个入度)。如果diff<0,返回false;如果节点非空,再令diff + 2(因为节点提供了2个出度)。

最后判断diff==0;

理解:如果在遍历过程中的某时刻,系统的入度>出度,则说明当前序列中出现过的所有分支节点的“空闲分支”均已用完,后序节点没有办法挂载到之前出现的节点之上,从而判定先序遍历的序列是不合法的。

class Solution {
public:
	bool isValidSerialization(string preorder) {
		stringstream ss(preorder);
		string val;
		int diff = 1;
		while (getline(ss, val, ',')){
			diff -= 1;
			if (diff < 0) return false;
			if (val != "#")
				diff += 2;
		}
		return diff == 0;
	}
};


【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.
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值