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 [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    3

Note:
Bonus points if you could solve it both recursively and iteratively.
題意：對(duì)稱(鏡像)二叉樹
思路：什么是對(duì)稱(鏡像)二叉樹?
1.兩個(gè)根結(jié)點(diǎn)的值要相等
2.左邊的左子樹和右邊的右子樹對(duì)稱
3.左邊的右子樹和右邊的左子樹對(duì)稱
注意:
1.空結(jié)點(diǎn)對(duì)稱
2.如果左子樹和右子樹只有一個(gè)為空則不對(duì)稱

/**
 * 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 dfs(root->left,root->right);
    }
    bool dfs(TreeNode*p,TreeNode*q)
    {
        if(!p||!q)
            return !p && !q;//如果左子樹和右子樹只有一個(gè)為空則不對(duì)稱
        return p->val==q->val&&dfs(p->left,q->right)&&dfs(p->right,q->left);
        
    }
};

迭代：

/**
 * 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;
        stack<TreeNode*>left,right;
        auto l=root->left,r=root->right;
        while(l||r||left.size()||right.size())
        {
            while(l&&r)
            {
                left.push(l),
                l=l->left;
                right.push(r);
                r=r->right;    
            }        
            if(l||r)
                return false;
            l=left.top(),left.pop();
            r=right.top(),right.pop();
            if(l->val!=r->val)
                return false;
            l=l->right,r=r->left;
        }
        return true;
    }
};