[leetcode-100-Same Tree]

Posted 2020-09-01 hellowOOOrld

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.

首先是递归版本:

bool isSameTree(TreeNode* p, TreeNode* q)
    {//递归
        if (p == NULL && q == NULL) return true;
        if (p == NULL && q != NULL || p != NULL && q == NULL || q->val != p->val) return false;

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

运行效率：

对比一下非递归版本：

bool isSameTree2(TreeNode* p, TreeNode* q)
    {//先序非递归
        stack<TreeNode*>st1, st2;
        if (p != NULL)st1.push(p);
        if (q!= NULL)st2.push(q);
        TreeNode* ptemp;
        TreeNode* qtemp;
        while (!st1.empty() && !st2.empty())
        {
            ptemp = st1.top();
            qtemp = st2.top();
            if (ptemp->val != qtemp->val) return false;
            st1.pop();
            st2.pop();
            if (ptemp->right != NULL) st1.push(ptemp->right);
            if (qtemp->right != NULL) st2.push(qtemp->right);
            if (st1.size() != st2.size()) return false;//比较两个栈的大小 

            if (ptemp->left != NULL) st1.push(ptemp->left);
            if (qtemp->left != NULL) st2.push(qtemp->left);
            if (st1.size() != st2.size()) return false;//比较两个栈的大小 
        }
        return (st1.size() == st2.size());
    }

运行效率：

可见，非递归确实效率要高一些。