leetcode101. Symmetric Tree

Q:

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.

思路

迭代法：類似於判斷兩個二叉樹是否相同，改變壓棧順序即可

AC解：

class Solution {
public:
    bool isSymmetric(TreeNode* root) 
    {
        stack<TreeNode*> s;
        if (root == nullptr)
            return true;

        TreeNode *p = root->left, *q = root->right;
        s.push(p);
        s.push(q);

        while (!s.empty())
        {
            q = s.top();    s.pop();
            p = s.top();    s.pop();

            if (q == nullptr && p == nullptr)   continue;
            if (q == nullptr || p == nullptr)   return false;
            if (q->val != p->val)               return false;
            //一個從左往右入棧，一個從右往左入棧
            s.push(p->left);
            s.push(q->right);
            s.push(p->right);
            s.push(q->left);
        }

        return true;
    }
};

遞歸法：也類似於判斷兩個二叉樹是否相同的解法，改變傳入的參數即可

AC解：

 //遞歸解法
class Solution {
public:
    bool isSymmetric(TreeNode* root) 
    {
        if (root == nullptr)
            return true;

        TreeNode *p = root->left, *q = root->right;

        return recurse(p, q);
    }

    bool recurse(TreeNode *p, TreeNode *q)
    {
        if (p == nullptr && q == nullptr)   return true;//終止條件
        if (p == nullptr || q == nullptr)   return false;   //剪枝

        return p->val == q->val && recurse(p->left,q->right)    //三方合併
                && recurse(p->right,q->left);
    }
};