【Leetcode】Symmetric Tree

题目：

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.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

解题思路：采用先序遍历的思路对根节点的两颗子树进行遍历，在遍历的时候需要注意，左子树的遍历过程中，先访左子树的左儿子，然后访问左子树的右儿子；右子树的遍历过程中，先访问右子树的右儿子，再访问右子树的左儿子。即镜像的对左右两颗子树进行比较。

递归代码：

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

迭代代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    bool isSymmetric(TreeNode *root) {        stack<TreeNode *> s;        if(root==nullptr)return true;        s.push(root->left);        s.push(root->right);        while(!s.empty()){            TreeNode *Right=s.top();s.pop();            TreeNode *Left=s.top();s.pop();                        if(Left==nullptr&&Right==nullptr)continue;            if(Left==nullptr||Right==nullptr)return false;                        if(Left->val==Right->val){                s.push(Left->left);                s.push(Right->right);                s.push(Left->right);                s.push(Right->left);            }else{                return false;            }               }        return true;    }};