101. Symmetric Tree

problem description

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.

algorithm thought

本质是树的层次遍历，判断每一层是否是对称的。层次遍历需要用到队列来辅助解决。得到每一层所有的值。最后遍历判断是否对称。但是对于这题，还有个技巧。对于左右子树，采取不同的策略。对于左子树，每次访问完之后，先push左及节点再push右节点。右子树，先push右再左。这样，每次从队列中拿两个节点出来，这两个节点就刚好都是对称节点。我们判断它们是否一样就行，然后继续运行。

/**
 * 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) {
        TreeNode *left,*right;
        if(root==NULL)
            return true;
        queue<TreeNode*> qu;
        qu.push(root->left);
        qu.push(root->right);
        while(!qu.empty()){
            left=qu.front();
            qu.pop();
            right=qu.front();
            qu.pop();
            if(left==NULL&&right==NULL)
                continue;
            if(left==NULL||right==NULL)
                return false;
            if(left->val!=right->val)
                return false;
            qu.push(left->left);
            qu.push(right->right);
            qu.push(left->right);
            qu.push(right->left);
        }
        return true;
    }
};

algorithm analysis

时间复杂度就是O(n),遍历一遍树。