【多次过】Lintcode 1360. Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

样例

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,#,3,#,3} is not:

    1
   / \
  2   2
   \   \
   3    3

解题思路1：

    递归。判断是否对称，可以先将其中一子树翻转（类似于Lintcode 175. 翻转二叉树），然后看两子树是否相等（类似于Lintcode 469. Same Tree）

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: root of the given tree
     * @return: whether it is a mirror of itself 
     */
    bool isSymmetric(TreeNode * root) 
    {
        // Write your code here
        if(root == NULL)
            return true;
        
        TreeNode * left = root->left;
        TreeNode * right = root->right;
        right = revert(right);
        
        return is_identical(left,right);
    }
    
    TreeNode * revert(TreeNode * root)
    {
        if(root == NULL)
            return NULL;
        
        revert(root->left);
        revert(root->right);
        swap(root->left , root->right);
        return root;
    }
    
    bool is_identical(TreeNode * root1 , TreeNode * root2)
    {
        if(root1 == NULL && root2 == NULL)
            return true;
        else if(root1 != NULL && root2 == NULL)
            return false;
        else if(root1 == NULL && root2 != NULL)
            return false;
        else
        {
            if(root1->val != root2->val)
                return false;
            else
                return is_identical(root1->left,root2->left) && is_identical(root1->right,root2->right);
        }
    }
};

解题思路2：

    循环，使用辅助数据结构--队列。类似于广度优先遍历。

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: root of the given tree
     * @return: whether it is a mirror of itself 
     */
    bool isSymmetric(TreeNode * root) 
    {
        // Write your code here
        if(root == NULL)
            return true;
        
        queue<TreeNode *> q1;
        queue<TreeNode *> q2;
        q1.push(root);
        q2.push(root);
        
        while(!q1.empty() && !q2.empty())
        {
            TreeNode * node1 = q1.front();
            q1.pop();
            
            TreeNode * node2 = q2.front();
            q2.pop();
            
            if(node1 == NULL && node2 == NULL)
                continue;
            if(node1 == NULL || node2 == NULL)//若一个为空，一个不为空，则false
                return false;
            if(node1->val != node2->val)
                return false;
            
            q1.push(node1->left);
            q1.push(node1->right);
            
            q2.push(node2->right);
            q2.push(node2->left);
        }
        
        return q1.empty() && q2.empty();
    }