[面试真题] LeetCode：Symmetric Tree

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

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

解法 1：非递归解法

使用双端队列dequeue记录对称的节点，依次从dequeue的头尾取出两个节点进行判断，两个节点的值以及节点左右孩子结构分别相同的情况下，将他们的孩子按顺序加入dequeue。再依次从dequeue的头尾取出两个节点进行判断，直到dequeue为空。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(NULL == root){
            return true;
        }
        if((root->left && !root->right) || (!root->left && root->right)){
            return false;
        }
        if(!root->left && !root->right){
            return true;
        }
        if(root->left->val != root->right->val){
            return false;
        }
        deque<TreeNode*> dq;
        dq.push_front(root->left);
        dq.push_back(root->right);
        while(dq.size()){
            TreeNode *l = dq.front();
            dq.pop_front();
            TreeNode *r = dq.back();
            dq.pop_back();
            if(l->val != r->val){
                return false;
            }
            if(l->left && r->right){
                dq.push_front(l->left);
                dq.push_back(r->right);
            }else if((l->left && !r->right) || (!l->left && r->right)){
                return false;
            }
            if(l->right && r->left){
                dq.push_front(l->right);
                dq.push_back(r->left);
            }else if((l->right && !r->left) || (!l->right && r->left)){
                return false;
            }
        }
        return true;
    }
};

Run Status: Accepted!

Program Runtime: 20 milli secs

Progress: 190/190 test cases passed.

 

解法 2：递归解法

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    bool isSymmetric(TreeNode *left, TreeNode *right) {
        if(left && right){
            if(left->val != right->val){
                return false;
            }
        }
        if((left && !right) || (!left && right)){
            return false;
        }
        if(!left && !right){
            return true;
        }
        if(left->left){
            if(!right->right){
                return false;
            }else{
                if(left->left->val != right->right->val){
                    return false;    
                }
            }
        }else{
            if(right->right){
                return false;
            }
        }
        if(left->right){
            if(!right->left){
                return false;
            }else{
                if(left->right->val != right->left->val){
                   return false; 
                }
            }
        }else{
            if(right->left){
                return false;
            }
        }
        return (left->left ? isSymmetric(left->left, right->right) : true) && (left->right ? isSymmetric(left->right, right->left) : true);
    }
        
public:
    bool isSymmetric(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(NULL == root){
            return true;
        }
        return isSymmetric(root->left, root->right);
    }
};

Run Status: Accepted!
Program Runtime: 24 milli secs

Progress: 190/190 test cases passed.