[LeetCode] Same Tree 判断相同树

2024年1月5日 363次阅读 来源: Grandyang

 

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.

 

判断两棵树是否相同和之前的判断两棵树是否对称都是一样的原理，利用深度优先搜索DFS来递归。代码如下：

 

解法一：

class Solution {
public:
    bool isSameTree(TreeNode *p, TreeNode *q) {
        if (!p && !q) return true;
        if ((p && !q) || (!p && q) || (p->val != q->val)) return false;
        return isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
    }
};

 

这道题还有非递归的解法，因为二叉树的四种遍历(层序，先序，中序，后序)均有各自的迭代和递归的写法，这里我们先来看先序的迭代写法，相当于同时遍历两个数，然后每个节点都进行比较，参见代码如下：

 

解法二：

class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        stack<TreeNode*> s1, s2;
        if (p) s1.push(p);
        if (q) s2.push(q);
        while (!s1.empty() && !s2.empty()) {
            TreeNode *t1 = s1.top(); s1.pop();
            TreeNode *t2 = s2.top(); s2.pop();
            if (t1->val != t2->val) return false;
            if (t1->left) s1.push(t1->left);
            if (t2->left) s2.push(t2->left);
            if (s1.size() != s2.size()) return false;
            if (t1->right) s1.push(t1->right);
            if (t2->right) s2.push(t2->right);
            if (s1.size() != s2.size()) return false;
        }
        return s1.size() == s2.size();
    }
};

 

其他几种遍历顺序的迭代写法应该也能实现，有时间补充完整~

 

参考资料：

https://leetcode.com/discuss/22197/my-non-recursive-method

 

    原文作者：Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/4053384.html
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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