[Leetcode] Same Tree

1最容易想到的递归调用

bool isSameTree(TreeNode *r1, TreeNode *r2)
{
    if(r1 == NULL && r2 == NULL) return true; //终止条件
    if(r1 == NULL || r2 == NULL) return false;//减枝，我觉得也是终止条件

    if(r1->val == r2->val
        && isSameTree(r1->left, r2->left)
        && isSameTree(r1->right, r2->right)) //递归，三方合并
        return true;
    return false;
}

2 迭代，受下面前序方法的启发

前序：

class Solution {
    public:
        vector<int> preorderTraversal(TreeNode *root) {
            vector<int> result;
            const TreeNode *p;
            stack<const TreeNode *> s;
            p = root;
            if (p != nullptr) s.push(p);
            while (!s.empty()) {
                p = s.top();
                s.pop();
                result.push_back(p->val);
                if (p->right != nullptr) s.push(p->right);
                if (p->left != nullptr) s.push(p->left);
            }
            return result;
        }
};

迭代：

class Solution {
    public:
        bool isSameTree(TreeNode *r1, TreeNode* r2) {

            if(r1 == NULL && r2 == NULL) return true;
            if(r1 == NULL || r2 == NULL) return false;

            stack<TreeNode*> stack;
            stack.push(r1);
            stack.push(r2);

            TreeNode *p1;
            TreeNode *p2;

            while(!stack.empty())
            {
                p1 = stack.top();
                stack.pop();
                p2 = stack.top();
                stack.pop();

                if(p1 == NULL && p2 == NULL) continue;//都为空时无法调用p1->right，p1->left
                if(p1 == NULL || p2 == NULL) return false;
                if(p1->val != p2->val) return false;

                stack.push(p1->right);
                stack.push(p2->right);

                stack.push(p1->left);
                stack.push(p2->left);
                
            }

            return true;
        }

};