101. Symmetric Tree

原创于 2018-01-29 08:17:51 发布 · 194 阅读

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文章标签：

#Tree

本文介绍了一种检查二叉树是否为自身镜像的方法，即判断二叉树是否围绕其中心对称。提供了递归和非递归两种解决方案，并通过示例展示了算法的应用。

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:

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

Note:
Bonus points if you could solve it both recursively and iteratively.

递归：

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        
        if(!root) return true;
        return helper(root->left, root->right);
    }

    bool helper(TreeNode *p, TreeNode *q)
    {
        if(!p && !q)
            return true;
        if(!p || !q)
            return false;
        if( p->val != q->val )
            return false;
        return helper(p->left,q->right) && helper(p->right, q->left);
    }
};

非递归：

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(!root) return true;
        if(!root->left&&!root->right) return true;
        if(!root->left&&root->right || root->left&&!root->right) return false;
        deque<TreeNode *> dq;
        dq.push_back(root->left);
        dq.push_back(root->right);
        while(!dq.empty())
        {
            TreeNode * lroot = dq.front();
            TreeNode * rroot = dq.back();
            dq.pop_front();
            dq.pop_back();
            if(lroot->val!=rroot->val) return false;
            if(!lroot->left&&rroot->right || lroot->left&&!rroot->right) return false;
            if(lroot->left)
            {
                dq.push_front(lroot->left);
                dq.push_back(rroot->right);
            }
            if(lroot->right&&!rroot->left || !lroot->right&&rroot->left) return false;
            if(lroot->right)
            {
                dq.push_front(lroot->right);
                dq.push_back(rroot->left);
            }
            
        }
        return true;
        
    }
};