【Leetcode】572. Subtree of Another Tree 解题报告

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求一棵树是不是另一棵树的子树

方法1 双递归

对母树的每个节点进行访问，判断以该节点为定点的树是不是和子树相同

class Solution1:
    def isSubtree(self, s, t):
        """
        :type s: TreeNode
        :type t: TreeNode
        :rtype: bool
        """
        if s == None:
            return False
        if self.isSame(s, t):
            return True
        else:
            return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)



    def isSame(self, rootA, rootB):
        if rootA == None and rootB == None:
            return True
        if rootA == None or rootB == None:
            return False
        else:
            if rootA.val != rootB.val:
                return False
            else:
                return self.isSame(rootA.left, rootB.left) and self.isSame(rootA.right, rootB.right)

方法2 方法1的非递归形式

class Solution2:
    def isSubtree(self, s, t):
        """
        :type s: TreeNode
        :type t: TreeNode
        :rtype: bool
        """
        if s == None:
            return False
        stack = [s]
        while(len(stack)):
            node = stack.pop()
            if self.isSame(node, t):
                return True
            else:
                if node.left:
                    stack.append(node.left)
                if node.right:
                    stack.append(node.right)
        return False

    def isSame(self, rootA, rootB):
        stack = [[rootA, rootB]]
        while(len(stack)):
            nodeA,nodeB = stack.pop()
            if nodeA == None and nodeB == None:
                continue
            else:
                if nodeA == None or nodeB == None:
                    if nodeA != nodeB:
                        return False
                else:
                    if nodeA.val != nodeB.val:
                        return False
                    else:
                        stack.append([nodeA.left, nodeB.left])
                        stack.append([nodeA.right, nodeB.right])
        return True

方法3 序列化法

求出母树和子树的字符串，看子树的字符串和是否在母树的字符串。
注：遍历二叉树的时候必须用先序遍历，而且每个节点要访问完其左右子节点为止（就算为None也要访问）

class Solution3:
    def isSubtree(self, s, t):
        """
        :type s: TreeNode
        :type t: TreeNode
        :rtype: bool
        """
        s_string = []
        t_string = []
        self.serialize(s, s_string)
        self.serialize(t, t_string)
        return "".join(t_string) in "".join(t_string)

    def serialize(self, root, path):
        if root == None:
            path.append(",#")
        else:
            path.append(",%s"%(str(root.val)))
            self.serialize(root.left, path)
            self.serialize(root.right, path)