daily leetcode - same-tree - !

题目地址

https://leetcode.com/problems/same-tree/

题目描述

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1
          / \       / \
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

Example 2:

Input:     1         1
          /           \
         2             2

        [1,2],     [1,null,2]

Output: false

Example 3:

Input:     1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false

思路

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。

关键点解析

代码

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,可参见之间那道 Binary Tree Preorder Traversal,参见代码如下:

解法二:

class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        stack<TreeNode*> st;
        st.push(p); st.push(q);
        while (!st.empty()) {
            p = st.top(); st.pop();
            q = st.top(); st.pop();
            if (!p && !q) continue;
            if ((p && !q) || (!p && q) || (p->val != q->val)) return false;
            st.push(p->right); st.push(q->right);
            st.push(p->left); st.push(q->left);
        }
        return true;
    }
};

也可以使用中序遍历的迭代写法,对应之前那道 Binary Tree Inorder Traversal,参见代码如下:

解法三:

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

对于后序遍历的迭代写法,貌似无法只是用一个栈来做,因为每次取出栈顶元素后不立马移除,这样使用一个栈的话两棵树结点的位置关系就会错乱,分别使用各自的栈就好了,对应之前那道 Binary Tree Postorder Traversal,参见代码如下:

解法四:

class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        stack<TreeNode*> st1, st2;
        TreeNode *head1, *head2;
        while (p || q || !st1.empty() || !st2.empty()) {
            while (p || q) {
                if ((p && !q) || (!p && q) || (p->val != q->val)) return false;
                st1.push(p); st2.push(q);
                p = p->left; q = q->left;
            }
            p = st1.top(); 
            q = st2.top(); 
            if ((!p->right || p->right == head1) && (!q->right || q->right == head2)) {
                st1.pop(); st2.pop();
                head1 = p; head2 = q;
                p = nullptr; q = nullptr;
            } else {
                p = p->right;
                q = q->right;
            }
        }
        return true;
    }
};

对于层序遍历的迭代写法,其实跟先序遍历的迭代写法非常的类似,只不过把栈换成了队列,对应之前那道 Binary Tree Level Order Traversal,参见代码如下:

解法五:

class Solution {
public:
    bool isSameTree(TreeNode* p, TreeNode* q) {
        queue<TreeNode*> que;
        que.push(p); que.push(q);
        while (!que.empty()) {
            p = que.front(); que.pop();
            q = que.front(); que.pop();
            if (!p && !q) continue;
            if ((p && !q) || (!p && q) || (p->val != q->val)) return false;
            que.push(p->right); que.push(q->right);
            que.push(p->left); que.push(q->left);
        }
        return true;
    }
};

本文参考自:
https://github.com/grandyang/leetcode/ &
https://github.com/azl397985856/leetcode


标题: daily leetcode - same-tree - !
文章作者: lonuslan
文章链接: https://louislan.com/articles/2020/04/19/1587263786203.html
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