LeetCode Same Tree

Same Tree Total Accepted: 52651 Total Submissions: 125166 My Submissions Question Solution
Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

题意：
比较两棵二叉树是否完全一致
递归版解法：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSameTree(TreeNode *p, TreeNode *q) {
        if(!p&&!q)
            return true;
        if(p&&!q)
            return false;
        if(!p&&q)
            return false;
        if(p->val!=q->val)
            return false;
        return isSameTree(p->left,q->left) && isSameTree(p->right,q->right);
    }
};

54 / 54 test cases passed.
Status: Accepted
Runtime: 2 ms

非递归版解法(网络整理)：

/**
 * 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 isSameTree(TreeNode *p, TreeNode *q) {
        if(!p && !q)
            return true;
        else if(!p && q)
            return false;
        else if(p && !q)
            return false;
        else
        {
            if(p->val != q->val)
                return false;
            else
            {    
                queue<TreeNode*> lq;
                queue<TreeNode*> rq;
                lq.push(p);
                rq.push(q);
                while(!lq.empty() && !rq.empty())
                {
                    TreeNode* lfront = lq.front();
                    TreeNode* rfront = rq.front();
                    lq.pop();
                    rq.pop();
                    if(!lfront->left && !rfront->left)
                        ;// null
                    else if(!lfront->left && rfront->left)
                        return false;
                    else if(lfront->left && !rfront->left)
                        return false;
                    else
                    {
                        if(lfront->left->val != rfront->left->val)
                            return false;
                        else
                        {
                            lq.push(lfront->left);
                            rq.push(rfront->left);
                        }
                    }
                    if(!lfront->right && !rfront->right)
                        ;// null
                    else if(!lfront->right && rfront->right)
                        return false;
                    else if(lfront->right && !rfront->right)
                        return false;
                    else
                    {
                        if(lfront->right->val != rfront->right->val)
                            return false;
                        else
                        {
                            lq.push(lfront->right);
                            rq.push(rfront->right);
                        }
                    }    
                }
                return true;
            }
        }
    }
};

54 / 54 test cases passed.
Status: Accepted
Runtime: 3 ms