100. Same Tree

最新推荐文章于 2024-10-14 18:27:49 发布

转载最新推荐文章于 2024-10-14 18:27:49 发布 · 260 阅读

本文介绍了两种判断二叉树是否相等的方法：递归解法及层序遍历法。递归解法通过比较根节点及其左右子树来判断；层序遍历法则借助队列保存节点并逐一比较。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

算法1：递归解法，如果根节点相同，再递归的看左子树和右子树是否相同

/** 
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(p && q)
        {
            if(p->val == q->val && isSameTree(p->left, q->left) && 
                isSameTree(p->right, q->right))
                return true;
            else return false;
        }
        else if(p || q)
            return false;
        else return true;
    }
};

算法2：通过层序遍历，分别用两个队列保存两棵树的层序节点。每次从两个队列中分别取出一个元素，如果他们的值相同，再继续遍历子节点。注意每次取出的两个元素的左子树要么同时非空，要么同时为空，右子树也一样

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        queue<TreeNode*> pQueue, qQueue;
        if(p)pQueue.push(p);
        if(q)qQueue.push(q);
        while(pQueue.empty() == false && qQueue.empty() == false)
        {
            TreeNode *pp = pQueue.front();
            TreeNode *qq = qQueue.front();
            pQueue.pop();  qQueue.pop();
            if(pp->val == qq->val)
            {
                if(pp->left && qq->left)
                {
                    pQueue.push(pp->left);
                    qQueue.push(qq->left);
                }
                else if(pp->left || qq->left)
                    return false;
                if(pp->right && qq->right)
                {
                    pQueue.push(pp->right);
                    qQueue.push(qq->right);
                }
                else if(pp->right || qq->right)
                    return false;
            }
            else return false;
        }
        if(pQueue.empty() && qQueue.empty())
            return true;
        else return false;
    }
};