LeetCode Symmetric Tree

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.

题意：给出一个二叉树，判断该树是不是对称的

思路：用DFS，首先左树与右树的结点开始值要相等，然后再将左树的左子树与右树的右子树比较，接着是左树的右子树与右树的左子树比较

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private boolean checkSymmetric(TreeNode left, TreeNode right)
    {
        if (left == null && right == null) return true;
        
        if (left == null || right == null) return false;
        
        if (left.val != right.val) return false;
        
        return checkSymmetric(left.left, right.right) && checkSymmetric(left.right, right.left);
    }
    
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        
        return checkSymmetric(root.left, root.right);
    }
}

第二种方法：用非递归方法，实际上就是模拟递归的入栈，出栈操作

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

};