LeetCode #100: Same Tree (C/C++).

easy

source: https://leetcode.com/problems/same-tree
C/C++ Solution to LeetCode problem 100. Same Tree.

Problem

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Given the roots of two binary trees p and q, 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.

Examples

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Example 1:

Input: p = [1,2,3], q = [1,2,3]
Output: true

Example 2:

Input: p = [1,2], q = [1,null,2]
Output: false

Example 3:

Input: p = [1,2,1], q = [1,1,2]
Output: false

Constraints

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• The number of nodes in both trees is in the range [0, 100].
• -104 <= Node.val <= 104

Solution

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We will explore both trees simultaneously with a BFS (Breadth First Search) algorithm.

• Insert the root nodes to the queue using pair<TreeNode*, TreeNode*>.
• While the queue has nodes:
  • If one node exists, but the other no, return false.
  • If both nodes exist:
    • If their values are different, return false.
    • Insert to the queue the pointers to the left and right of each node.
  • Pop the first element (pair of nodes) from the queue
• Return true.

---

/**
 * 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 isSameTree(TreeNode* p, TreeNode* q) {
    queue<pair<TreeNode *, TreeNode *>> bfs;

    bfs.push({p, q});
    TreeNode *a;
    TreeNode *b;
    
    while (!bfs.empty()) {
      a = bfs.front().first;
      b = bfs.front().second;
      if ((!a && b) || (a && !b))
        return false;
      if (a && b) {
        if (a->val != b->val)
          return false;

        bfs.push({a->left, b->left});
        bfs.push({a->right, b->right});
      }
      
      bfs.pop();
    }

    return true;
  }
};