判断是否是相同二叉树&&判断是否是对称树

今天药忘吃喽~ 2022-06-09 06:09 243阅读 0赞

LeetCode 100. Same Tree

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.

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

写递归先写最底层，将最底层情况写出，然后再根据逻辑，写出上层到底层的传递方式。

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Leetcode 101. Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree `[1,2,2,3,4,4,3]` is symmetric:

1
       / \
      2   2
     / \ / \
    3  4 4  3

But the following `[1,2,2,null,3,null,3]` is not:

1
       / \
      2   2
       \   \
       3    3

Note:  
Bonus points if you could solve it both recursively and iteratively.

提示给出可以用递归和非递归两种方式，先用递归写。

class Solution {
    public:
        bool isSymmetric(TreeNode* root) {
          //  TreeNode *cur = root;
            if (root == NULL)
                return true;
            return isMirror(root->left, root->right);
        }
        bool isMirror(TreeNode *p, TreeNode *q)
        {
            if (q == NULL && p == NULL)
                return true;
            if (p == NULL || q == NULL)
                return false;
            if (p->val == q->val)
                return isMirror(p->right, q->left) && isMirror(p->left, q->right);
            return false;
        }
    };

在写递归过程中，由于必须分叉，因此增加了一个带2参数的函数来辅助判断。

对于非递归，将二叉树分成两半，则此时思路就是对于左半部分按照从左向右层序遍历；对于右半部分从右向左进行层序遍历。层序遍历采用队列结构，因此构造两个队列。

/**
     * Definition for a binary tree node.
     * struct TreeNode {
     *     int val;
     *     TreeNode *left;
     *     TreeNode *right;
     *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     * };
     */
    class Solution {
    public:
        bool isSymmetric(TreeNode* root) {
          TreeNode *lchild;
          TreeNode *rchild;
          queue<TreeNode *>p, q;
          if (root == NULL) 
          {
              return true;
          }
          p.push(root->left);
          q.push(root->right);
            
         while (!p.empty() && !q.empty())
         {
             lchild = p.front();
             p.pop();
             rchild = q.front();
             q.pop();
             if (lchild == NULL && rchild == NULL)
                 continue;
             if (lchild == NULL || rchild == NULL)
                 return false;
             if (lchild->val != rchild->val)
                 return false;
             p.push(lchild->left);
             p.push(lchild->right);
             q.push(rchild->right);
             q.push(rchild->left);
         }
            return true;
        }
    };

需要注意的是continue的使用，因为此时队列尚不为空，因此并不能直接return，需要继续判断直到队列为空。