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  • 【LeetCode】100. Same Tree (2 solutions)

    Same Tree

    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.

    解法一:递归

    /**
     * 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
                    return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
            }
        }
    };

    解法二:非递归

    建立两个队列分别进行层次遍历,进队时检查对应点是否相等

    /**
     * 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(!isSameNode(p, q))
                return false;
            if(!p && !q)
                return true;
            
            queue<TreeNode*> lqueue;
            queue<TreeNode*> rqueue;
            lqueue.push(p);
            rqueue.push(q);
            while(!lqueue.empty() && !rqueue.empty())
            {
                TreeNode* lfront = lqueue.front();
                TreeNode* rfront = rqueue.front();
    
                lqueue.pop();
                rqueue.pop();
                
                if(!isSameNode(lfront->left, rfront->left))
                    return false;
                if(lfront->left && rfront->left)
                {
                    lqueue.push(lfront->left);
                    rqueue.push(rfront->left);
                }
                
                if(!isSameNode(lfront->right, rfront->right))
                    return false;
                if(lfront->right && rfront->right)
                {
                    lqueue.push(lfront->right);
                    rqueue.push(rfront->right);
                }
            }
            return true;
        }
        bool isSameNode(TreeNode* p, TreeNode *q)
        {
            if(!p && !q)
                return true;
            if((p && !q) || (!p && q) || (p->val != q->val))
                return false;
            return true;
        }
    };

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  • 原文地址:https://www.cnblogs.com/ganganloveu/p/4136259.html
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