Verify Preorder Serialization of a Binary Tree
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对于二叉树,我们把空的地方也作为叶子节点(如题目中的#),那么有
- 所有的非空节点提供2个出度和1个入度(根除外)
- 所有的空节点但提供0个出度和1个入度
我们在遍历的时候,计算diff = outdegree – indegree. 当一个节点出现的时候,diff – 1,因为它提供一个入度;当节点不是#的时候,diff+2(提供两个出度) 如果序列式合法的,那么遍历过程中diff >=0 且最后结果为0.
public class Solution { public boolean isValidSerialization(String preorder) { String[] strs = preorder.split(","); int diff = 1; for (String str: strs) { if (--diff < 0) { return false; } if (!str.equals("#")) { diff = diff + 2; } } return diff == 0; }}
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- Verify Preorder Serialization of a Binary Tree
- Verify Preorder Serialization of a Binary Tree
- Verify Preorder Serialization of a Binary Tree
- Verify Preorder Serialization of a Binary Tree
- Verify Preorder Serialization of a Binary Tree
- Verify Preorder Serialization of a Binary Tree
- Verify Preorder Serialization of a Binary Tree
- Verify Preorder Serialization of a Binary Tree
- Verify Preorder Serialization of a Binary Tree
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