leetcode_951. Flip Equivalent Binary Trees_二叉树遍历

https://leetcode.com/problems/flip-equivalent-binary-trees/

判断两棵二叉树是否等价：若两棵二叉树可以通过任意次的交换任意节点的左右子树变为相同，则称两棵二叉树等价。

思路：遍历二叉树，判断所有的子树是否等价。

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x)
        : val(x), left(NULL), right(NULL) {}
};

class Solution
{
public:
    void exchangeSons(TreeNode* root)
    {
        TreeNode* tmp = root->left;
        root->left = root->right;
        root->right = tmp;
    }

    int getNum(TreeNode* node)
    {
        if(node == NULL)
            return -1;
        else
            return node->val;
    }
    int compareSons(TreeNode* root1, TreeNode* root2)
    {
        TreeNode* left1 = root1->left;
        TreeNode* right1 = root1->right;
        TreeNode* left2 = root2->left;
        TreeNode* right2 = root2->right;
        int l1,l2,r1,r2;
        l1 = getNum(left1);
        l2 = getNum(left2);
        r1 = getNum(right1);
        r2 = getNum(right2);
        if(l1 == l2 && r1 == r2)
            return 1;
        else if(l1 == r2 && r1 == l2)
            return 2;
        else
            return 0;
    }
    bool flipEquiv(TreeNode* root1, TreeNode* root2)
    {
        if(root1 == NULL && root2 == NULL)
            return 1;
        else if(root1 == NULL)
            return 0;
        else if(root2 == NULL)
            return 0;
        int comres = compareSons(root1, root2);
        if(comres == 0)
            return 0;
        else if(comres == 2)
            exchangeSons(root2);
        bool leftEquiv = 1,rightEquiv = 1;
        if(root1->left != NULL)
            leftEquiv = flipEquiv(root1->left, root2->left);
        if(root1->right != NULL)
            rightEquiv = flipEquiv(root1->right, root2->right);
        if(leftEquiv&&rightEquiv)
            return 1;
        else
            return 0;
    }
};

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:

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