LeetCode331. Verify Preorder Serialization of a Binary Tree

最新推荐文章于 2024-10-03 19:13:18 发布

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最新推荐文章于 2024-10-03 19:13:18 发布

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本文探讨了如何在不重建二叉树的情况下，验证给定的字符串是否为正确的二叉树前序遍历序列化。通过分析二叉树的结构特性，提出了一种算法来检查序列化的有效性。

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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:

Input: "9,3,4,#,#,1,#,#,2,#,6,#,#"
Output: true

Example 2:

Input: "1,#"
Output: false

Example 3:

Input: "9,#,#,1"
Output: false

解决这道题需要观察一颗二叉树的结构。

如果一颗二叉树的非叶子节点（题目中非‘#’）个数为node，那么叶子节点（题目中的‘#’）则是node+1。

利用上述发现，我们可以这样判断输入是否合法：

1 叶子节点=非叶子节点+1，则说明已经是一颗完整的二叉树。那后面还有节点，不管是叶子节点还是非叶子节点都是非法的。

2 永远不会出现：叶子节点>非叶子节点+1 的情况。

class Solution {
public:
	bool isValidSerialization(string preorder) {
		int node(0), leaf(0), i(0);
		for (; i < preorder.size(); i++) {
			if (preorder[i] == '#') {
				leaf++;
			}
			else if (preorder[i] == ',') {
				continue;
			}
			else {
				node++;
				while (i < preorder.size() && isdigit(preorder[i])) {
					i++;
				}
				i--;
			}
			if (leaf > node + 1) return false;
			if (leaf == node + 1){
                i++;
                break;
            }
		}
		while (i < preorder.size()) {
			if (preorder[i] == '#' || isdigit(preorder[i])) return false;
			i++;
		}
		return leaf == node + 1;
	}
};