572. 是否为另一棵二叉树的子树 Subtree of Another Tree

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node‘s descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

     3
    /    4   5
  /  1   2

Given tree t:

   4
  /  1   2

Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

     3
    /    4   5
  /  1   2
    /
   0

Given tree t:

   4
  /  1   2

Return false.

题意：判断一棵二叉树是否为另一棵二叉树的子树

解法：先序遍历二叉树，并生成字符串（用#表示孩子节点是否为null的情况），判断字符串的包含关系

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public bool IsSubtree(TreeNode s, TreeNode t) {
    	string s1 = Tree2String(s);
    	string s2 = Tree2String(t);
    
    	return s1.Contains(s2) || s2.Contains(s1);
    }
    
	static public string Tree2String(TreeNode root) {
		if (root == null) {
			return "";
		}
		StringBuilder sb = new StringBuilder();

		Stack<TreeNode> stack = new Stack<TreeNode>();
		TreeNode current = root;

		stack.Push(current);

		while (stack.Count != 0) {
			TreeNode popelem = stack.Pop();
			/**
			 * 用#表示孩子节点是否为null的情况
			 * 用","号分隔节点是防止"12##"和"2##"这种情况
			 * ",12##",",2##"
			 */
			if (popelem == null) {
				sb.Append("#");
			} else {
				sb.Append(popelem.val);
			}
			if (popelem != null) {
				stack.Push(popelem.right);
				stack.Push(popelem.left);
			}
		}

		return sb.ToString();
	}
}

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