951. Flip Equivalent Binary Trees

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My Solution

The Key Idea for Solving This Coding Question

DFS (recursion) preorder traversal.
If both subtrees are not empty, visit the subtree whose root value is smaller.

C++ Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

class Solution {
public:
    bool flipEquiv(TreeNode *root1, TreeNode *root2) {
        vector<int> traversal1, traversal2;

        dfs(root1, traversal1);
        dfs(root2, traversal2);

        return traversal1 == traversal2;
    }

private:
    void dfs(TreeNode *root, vector<int> &traversal) {
        if (root == nullptr) {
            traversal.push_back(-1);
            return;
        }

        traversal.push_back(root->val);
        if (root->left == nullptr) {
            dfs(root->right, traversal);
            traversal.push_back(-1);
            return;
        }

        if (root->right == nullptr) {
            dfs(root->left, traversal);
            traversal.push_back(-1);
            return;
        }

        if (root->left->val < root->right->val) {
            dfs(root->left, traversal);
            dfs(root->right, traversal);
            return;
        }

        dfs(root->right, traversal);
        dfs(root->left, traversal);
    }
};

Time Complexity

$O(n)$

$n$ is the number of nodes in the binary tree.

Space Complexity

$O(H)$

$H$ is the height of the binary tree.

Miscellaneous