Leetcode -- Verify Preorder Serialization of a Binary Tree

题目描述

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

解答：

解法一：

关键思想是利用先序遍历的特征。首先是根节点然后左子树右子树，“#”代指空节点，依据这种遍历的特征，我们可以不断删除叶子节点。

利用栈结构，不断删除叶子节点，例如遇到“4 # #”的时候删除并换成“#”，直到最后不能再删除。例如图示。

	public static boolean isValidSerialization(String preorder) {
		List<String> stack = new ArrayList<String>();
		String[] arr = preorder.split(",");
		for (int i = 0; i < arr.length; i++) {
			stack.add(arr[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;
	}

解法二：

把二叉树的空节点当作叶子节点#表示。所有的叶子节点提供1个入度0个出度，非叶子节点2个出度1个入度（根除外），又二叉树的性质叶子节点个数=非叶子个数+1. 所以我们可以一个一个的遍历，计算diff = outdegree –indegree。对于内部节点，入度减一出度加二，叶子节点入度减一，如是正确序列结果diff必为0.

	public static boolean isValidSerialization(String preorder) {
		String[] nodes = preorder.split(",");
		int diff = 1;
		for (String node : nodes) {
			// 入度减一
			if (--diff < 0) return false;
			// 出度加二
			if (!node.equals("#")) diff += 2;
		}
		return diff == 0;
	}

refer：

https://discuss.leetcode.com/topic/35976/7-lines-easy-java-solution