LeetCode——Symmetric Tree

LeetCode——Symmetric Tree

# 101

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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

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

But the following [1,2,2,null,3,null,3] is not:

    1   / \  2   2   \   \   3    3Note:Bonus points if you could solve it both recursively and iteratively. 

这题是要求判断二叉树是否对称。要求用递归和迭代。这题的思路其实并不难，因为是二叉树。先写出以下几种情况：

1. root为空，返回true,开始递归2. root非空，无子节点，返回true3. 如果不是左右子节点都存在且相等，返回false4. 左右子节点存在且相当，再分别比较左右子树，形成递归

需要第一一个递归函数。这题的递归不难，只要把思路理清，很容易就写出来了。

• C++/递归方法

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

迭代的方法其实核心思想是一样的，只需要转换一下就可以了。

• C++/迭代方法

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isSymmetric(TreeNode* root) {        if(!root)            return true;        queue<TreeNode*>q1,q2;        q1.push(root -> left);        q2.push(root -> right);       while (!q1.empty() && !q2.empty()){            TreeNode* left = q1.front();            q1.pop();            TreeNode* right = q2.front();            q2.pop();            if (left == NULL && right == NULL)                continue;            if (left == NULL || right == NULL)                return false;            if (left -> val != right -> val)                return false;            q1.push(left ->l eft);            q1.push(left -> right);            q2.push(right -> right);            q2.push(right -> left);        }        return true;    }};

使用的是DFS深度优先搜索的算法，其实也可以用BFS广度优先搜索，差别不是很大，不同的比较方式，下面用Python实现。

• Python

# 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        queuel, queuer = [root.left], [root.right]        while len(queuel) > 0 and len(queuer) > 0:            left = queuel.pop()            right = queuer.pop()            if not left and not right:                continue            elif not left or not right:                return False            if left.val != right.val:                return False            queuel.insert(0, left.left)            queuel.insert(0, left.right)            queuer.insert(0, right.right)            queuer.insert(0, right.left)        return len(queuel) == 0 and len(queuer) == 0