Print the longest leaf to leaf path in a Binary tree

The diameter of a tree (sometimes called the width) is the number of nodes on the longest path between two end nodes. In this post, we will see how to print the nodes involved in the diameter of the tree. The diagram below shows two trees each with diameter nine, the leaves that form the ends of the longest path are shaded (note that there is more than one path in each tree of length nine, but no path longer than nine nodes).

Examples:

```Input:      1
/   \
2      3
/  \
4     5

Output : 4 2 1 3
or 5 2 1 3

Input:      1
/   \
2      3
/  \      \
4     5      6

Output : 4 2 1 3 6
or 5 2 1 3 6
```

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

We have already discussed how to find the diameter of a binary tree.Diameter of a Binary tree

We know that Diameter of a tree can be calculated by only using the height function because the diameter of a tree is nothing but the maximum value of (left_height + right_height + 1) for each node.
Now for the node which has the maximum value of (left_height + right_height + 1), we find the longest root to leaf path on the left side and similarly on the right side. Finally, we print left side path, root and right side path.
Time Complexity is O(N). N is the number of nodes in the tree.

 `// CPP program to print the longest leaf to leaf ` `// path ` `#include ` `using` `namespace` `std; ` ` `  `// Tree node structure used in the program ` `struct` `Node { ` `    ``int` `data; ` `    ``Node *left, *right; ` `}; ` ` `  `struct` `Node* newNode(``int` `data) ` `{ ` `    ``struct` `Node* node = ``new` `Node; ` `    ``node->data = data; ` `    ``node->left = node->right = NULL; ` ` `  `    ``return` `(node); ` `} ` ` `  `// Function to find height of a tree ` `int` `height(Node* root, ``int``& ans, Node*(&k), ``int``& lh, ``int``& rh,  ` `                                                     ``int``& f) ` `{ ` `    ``if` `(root == NULL) ` `        ``return` `0; ` ` `  `    ``int` `left_height = height(root->left, ans, k, lh, rh, f); ` ` `  `    ``int` `right_height = height(root->right, ans, k, lh, rh, f); ` ` `  `    ``// update the answer, because diameter of a ` `    ``// tree is nothing but maximum value of ` `    ``// (left_height + right_height + 1) for each node ` ` `  `    ``if` `(ans < 1 + left_height + right_height) { ` ` `  `        ``ans = 1 + left_height + right_height; ` ` `  `        ``// save the root, this will help us finding the ` `        ``//  left and the right part of the diameter ` `        ``k = root; ` ` `  `        ``// save the height of left & right subtree as well. ` `        ``lh = left_height; ` `        ``rh = right_height; ` `    ``} ` ` `  `    ``return` `1 + max(left_height, right_height); ` `} ` ` `  `// prints the root to leaf path ` `void` `printArray(``int` `ints[], ``int` `len, ``int` `f) ` `{ ` `    ``int` `i; ` ` `  `    ``// print left part of the path in reverse order ` `    ``if` `(f == 0) { ` `        ``for` `(i = len - 1; i >= 0; i--) { ` `            ``printf``(``"%d "``, ints[i]); ` `        ``} ` `    ``} ` ` `  `    ``// print right part of the path ` `    ``else` `if` `(f == 1) { ` `        ``for` `(i = 0; i < len; i++) { ` `            ``printf``(``"%d "``, ints[i]); ` `        ``} ` `    ``} ` `} ` ` `  `// this function finds out all the root to leaf paths ` `void` `printPathsRecur(Node* node, ``int` `path[], ``int` `pathLen,  ` `                                         ``int` `max, ``int``& f) ` `{ ` `    ``if` `(node == NULL) ` `        ``return``; ` ` `  `    ``// append this node to the path array ` `    ``path[pathLen] = node->data; ` `    ``pathLen++; ` ` `  `    ``// If it's a leaf, so print the path that led to here ` `    ``if` `(node->left == NULL && node->right == NULL) { ` ` `  `        ``// print only one path which is equal to the  ` `        ``// height of the tree. ` `        ``if` `(pathLen == max && (f == 0 || f == 1)) { ` `            ``printArray(path, pathLen, f); ` `            ``f = 2; ` `        ``} ` `    ``} ` ` `  `    ``else` `{ ` ` `  `        ``// otherwise try both subtrees ` `        ``printPathsRecur(node->left, path, pathLen, max, f); ` `        ``printPathsRecur(node->right, path, pathLen, max, f); ` `    ``} ` `} ` ` `  `// Computes the diameter of a binary tree with given root. ` `void` `diameter(Node* root) ` `{ ` `    ``if` `(root == NULL) ` `        ``return``; ` ` `  `    ``// lh will store height of left subtree ` `    ``// rh will store height of right subtree ` `    ``int` `ans = INT_MIN, lh = 0, rh = 0; ` ` `  `    ``// f is a flag whose value helps in printing ` `    ``// left & right part of the diameter only once ` `    ``int` `f = 0; ` `    ``Node* k; ` `    ``int` `height_of_tree = height(root, ans, k, lh, rh, f); ` `    ``int` `lPath[100], pathlen = 0; ` ` `  `    ``// print the left part of the diameter ` `    ``printPathsRecur(k->left, lPath, pathlen, lh, f); ` `    ``printf``(``"%d "``, k->data); ` `    ``int` `rPath[100]; ` `    ``f = 1; ` ` `  `    ``// print the right part of the diameter ` `    ``printPathsRecur(k->right, rPath, pathlen, rh, f); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// Enter the binary tree ... ` `    ``//           1 ` `    ``//         /   \      ` `    ``//        2     3 ` `    ``//      /   \    ` `    ``//     4     5 ` `    ``//      \   / \  ` `    ``//       8 6   7 ` `    ``//      / ` `    ``//     9 ` `    ``struct` `Node* root = newNode(1); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(3); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(5); ` `    ``root->left->right->left = newNode(6); ` `    ``root->left->right->right = newNode(7); ` `    ``root->left->left->right = newNode(8); ` `    ``root->left->left->right->left = newNode(9); ` ` `  `    ``diameter(root); ` ` `  `    ``return` `0; ` `} `

Output:

```9 8 4 2 5 6
```

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