LeetCode #951. Flip Equivalent Binary Trees

最新推荐文章于 2021-08-02 19:37:13 发布

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本文介绍了一种算法，用于确定两棵二叉树是否通过一系列节点子树左右交换操作后可以相等，即判断二叉树的翻转等价性。通过递归比较两个根节点及其子树来实现这一目标。

题目描述：

For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.

A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.

Write a function that determines whether two binary trees are flip equivalent. The trees are given by root nodes root1 and root2.

Example 1:

Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
Output: true
Explanation: We flipped at nodes with values 1, 3, and 5.

Note:

Each tree will have at most 100 nodes.
Each value in each tree will be a unique integer in the range [0, 99].

class Solution {
public:
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        if(root1==NULL&&root2==NULL) return true;
        else if(root1==NULL&&root2!=NULL) return false;
        else if(root1!=NULL&&root2==NULL) return false;
        else if(root1->val!=root2->val) return false;
        return (flipEquiv(root1->left,root2->left)&&flipEquiv(root1->right,root2->right)||
                flipEquiv(root1->left,root2->right)&&flipEquiv(root1->right,root2->left));
    }
};