Update code solution

Author: SartHak-0-Sach

200-18th_July-Number_of_good_leaf_node_pairs/approach-explanation.txt

@@ -0,0 +1,8 @@
+Intuition
+The idea is to first get the path to each leaf node.
+Then, compare each leaf path with every other leaf path to calculate the distance between them.
+Complexity
+Time complexity: O(n 
+2
+ ⋅logn)
+Space complexity: O(n⋅h)

200-18th_July-Number_of_good_leaf_node_pairs/code.cpp

@@ -0,0 +1,59 @@
+/**
+ * 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:
+    void getPath(TreeNode *root, vector<string> &path, string &st)
+    {
+        if (!root)
+            return;
+        if (!root->left && !root->right)
+            path.push_back(st);
+        if (root->left)
+        {
+            st += 'l';
+            getPath(root->left, path, st);
+            st.pop_back();
+        }
+        if (root->right)
+        {
+            st += 'r';
+            getPath(root->right, path, st);
+            st.pop_back();
+        }
+    }
+    int countPairs(TreeNode *root, int distance)
+    {
+        int ans = 0;
+        vector<string> path;
+        string st = "";
+        getPath(root, path, st);
+        for (int i = 0; i < path.size(); i++)
+        {
+            for (int j = i + 1; j < path.size(); j++)
+            {
+                int k = 0;
+                int tmp = 0;
+                while (k < path[i].size() && k < path[j].size())
+                {
+                    if (path[i][k] != path[j][k])
+                        break;
+                    k++;
+                }
+                tmp += path[i].size() - k + path[j].size() - k;
+                if (tmp <= distance)
+                    ans++;
+            }
+        }
+        return ans;
+    }
+};