leetcode101題 題解 翻譯 C語言版 Python版__Python

101. Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1   / \  2   2 / \ / \3  4 4  3

But the following is not:

    1   / \  2   2   \   \   3    3

Note:
Bonus points if you could solve it both recursively and iteratively. 101.對稱樹 給定一棵二叉樹，檢查它是否是它自身的鏡像（即：軸對稱） 舉個例子，這棵二叉樹是對稱的：

    1   / \  2   2 / \ / \3  4 4  3

但是下面這個就不是：

    1   / \  2   2   \   \   3    3

思路：這題用遞迴很好解決，開始時將左子樹和右子樹分別遞迴下去，如果第一棵樹的左子樹和第二棵樹的右子樹遞迴下去完全相同，同時第一棵樹的右子樹和第二棵樹的左子樹遞迴下去完全相同，那麼這兩棵樹就滿足對稱了。如果根結點的兩棵子樹滿足對稱了，那麼整棵樹就是對稱的了。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     struct TreeNode *left; *     struct TreeNode *right; * }; */  bool isSymmetriclr(struct TreeNode* left, struct TreeNode* right){     if (left == NULL && right == NULL) return true;     if (left == NULL && right != NULL) return false;     if (left != NULL && right == NULL) return false;     if (left->val != right->val) return false;     if (!isSymmetriclr(left->left, right->right)) return false;     if (!isSymmetriclr(left->right, right->left)) return false;     return true; } bool isSymmetric(struct TreeNode* root) {    if (root == NULL) return true;    return isSymmetriclr(root->left, root->right);}

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def isSymmetric(self, root):        """        :type root: TreeNode        :rtype: bool        """        if not root: return True        return self.isSymmetriclr(root.left, root.right)            def isSymmetriclr(self, left, right):        if not (left or right): return True        if (left and right==None) or (left==None and right):            return False        if left.val != right.val: return False        if not self.isSymmetriclr(left.left, right.right): return False        if not self.isSymmetriclr(left.right, right.left): return False        return True