Finding the lexicographically smallest diameter in a binary tree

• Difficulty Level : Expert
• Last Updated : 18 Jun, 2020

Given a binary tree where node values are lowercase alphabets, the task is to find the lexicographically smallest diameter. Diameter is the longest path between any two leaf nodes, hence, there can be multiple diameters in a Binary Tree. The task is to print the lexicographically smallest diameter among all possible diameters.

Examples:

Input:
a
/   \
b       c
/  \     /  \
d   e    f    g
Output: Diameter: 5
Lexicographically smallest diameter: d b a c f
Explanation:
Note that there are many other paths
exist like {d, b, a, c, g},
{e, b, a, c, f} and {e, b, a, c, g}
but {d, b, a, c, f}
is lexicographically smallest

Input:
k
/   \
e       s
/  \
g   f
Output: Diameter: 4
Lexicographically smallest diameter: f e k s
Explanation:
Note that many other paths
exist like {g, e, k, s}
{s, k, e, g} and {s, k, e, f}
but {f, e, k, s} is
lexicographically smallest

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
The approach is similar to finding diameter as discussed in the previous post. Now comes the part of
printing the longest path with the maximum diameter and lexicographically smallest.

Steps:

• Custom compare function returns lexicographical smallest vector is made.
• Six kinds of vector are been maintained which contains
the left subtree ( of a node) nodes in leftdiameter
the right subtree (of a node) nodes in rightdiameter
nodes occurring in the left height (of a node)
nodes occurring in the right height (of a node)
heightv vector contains the nodes occurring the path of max height
dia vector contains the nodes occurring the path of max height
• Rest part is explained in the comments of code and it will be difficult to explain here in words

Below is the implementation of the above approach:

 // C++ program for the above approach #include using namespace std;    // Binary Tree Node struct node {     char data;     node *left, *right; };    // Utility function to create a new node node* newNode(char data) {     node* c = new node;     c->data = data;     c->left = c->right = NULL;     return c; }    // Function to compare and return // lexicographically smallest vector vector compare(vector a, vector b) {     for (int i = 0; i < a.size() && i < b.size(); i++) {         if (a[i]->data < b[i]->data) {             return a;         }         if (a[i]->data > b[i]->data) {             return b;         }     }        return a; }    // Function to find diameter int diameter(node* root, int& height, vector& dia,              vector& heightv) {     // If root is null     if (!root) {         height = 0;         return 0;     }        // Left height and right height     // respectively     int lh = 0, rh = 0;        // Left tree diameter and     // right tree diameter     int ld, rd;        vector leftdia;     vector rightdia;     vector leftheight;     vector rightheight;        // Left subtree diameter     ld = diameter(root->left, lh, leftdia, leftheight);        // Right subtree diameter     rd = diameter(root->right, rh, rightdia, rightheight);        // If left height is more     // than right tree height     if (lh > rh) {            // Add current root so lh + 1         height = lh + 1;            // Change vector heightv to leftheight         heightv = leftheight;            // Insert current root in the path         heightv.push_back(root);     }        // If right height is     // more than left tree height     else if (rh > lh) {            // Add current root so rh + 1         height = rh + 1;            // Change vector heightv to rightheight         heightv = rightheight;            // Insert current root in the path         heightv.push_back(root);     }        // Both height same compare     // lexicographically now     else {            // Add current root so rh + 1         height = rh + 1;            // Lexigcographical comparison between two vectors         heightv = compare(leftheight, rightheight);            // Insert current root in the path         heightv.push_back(root);     }        // If distance of one leaf node to another leaf     // containing the root is more than the left     // diameter and right diameter     if (lh + rh + 1 > max(ld, rd)) {            // Make dia equal to leftheight         dia = leftheight;            // Add current root into it         dia.push_back(root);            for (int j = rightheight.size() - 1; j >= 0; j--) {             // Add right tree (right to root) nodes             dia.push_back(rightheight[j]);         }     }        // If either leftdiameter containing the left     // subtree and root or rightdiameter containg     // the right subtree and root is more than     // above lh+rh+1     else {            // If diameter of left tree is         // greater our answer vector i.e         // dia is equal to leftdia then         if (ld > rd) {             dia = leftdia;         }            // If both diameter         // same check lexicographically         else if (ld == rd) {             dia = compare(leftdia, rightdia);         }            // If diameter of right tree         // is greater our answer vector         // i.e dia is equal to rightdia then         else {             dia = rightdia;         }     }        return dia.size(); }    // Driver code int main() {     node* root = newNode('a');     root->left = newNode('b');     root->right = newNode('c');     root->left->left = newNode('d');     root->left->right = newNode('e');     root->right->left = newNode('f');     root->right->right = newNode('g');        int height = 0;     vector dia, heigh;     cout << "Diameter is: " << diameter(root, height,                                         dia, heigh)          << endl;        // Printing the lexicographically smallest diameter     cout << "Lexicographically smallest diameter:" << endl;     for (int j = 0; j < dia.size(); j++) {         cout << dia[j]->data << " ";     }        return 0; }

Output:

Diameter is: 5
Lexicographically smallest diameter:
d b a c f

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