Binary Tree Paths LeetCode (All Root to Leaf Path)

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

["1->2->5", "1->3"]

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Solution Source: Java

	import java.util.ArrayList;
	import java.util.List;
	/**
	*
	* Given a binary tree, return all root-to-leaf paths.
	*
	* For example, given the following binary tree:
	*
	* 1
	* / \
	* 2 3
	* \
	* 5
	* All root-to-leaf paths are:
	*
	* ["1->2->5", "1->3"]
	*
	*
	* Solution 1:
	* 1)start from the root node and traverse down its child
	* 2)take one branch at a time for every traversed node and add the value of the node to
	* the path string
	* 3)when the next child node (on left side and right side) is null, the string is complete,
	* else we just keep on appending the value of this node to the current path
	*
	*
	* @author rbaral
	*/
	/**
	* Definition for a binary tree node. public class TreeNode { int val; TreeNode
	* left; TreeNode right; TreeNode(int x) { val = x; } }
	*/
	public class BinaryTreePath {
	public static List<String> binaryTreePaths(TreeNode root) {
	List<String> pathList = new ArrayList<String>();
	if (root != null) {
	performDFS(root, "", pathList);
	}
	return pathList;
	}
	static void performDFS(TreeNode root, String path, List<String> pathList) {
	if (root.left == null && root.right == null) {
	pathList.add(path + root.val);
	}
	if (root.left != null) {
	performDFS(root.left, path + root.val + "->", pathList);
	}
	if (root.right != null) {
	performDFS(root.right, path + root.val + "->", pathList);
	}
	}
	public static void main(String args[]) {
	}
	}

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