Introduction to Root to Leaf Path Pattern

The Root to Leaf Path Pattern is fundamental when dealing with binary trees. This pattern focuses on problems that involve navigating from the root of the tree to its leaves, where each leaf represents a node without children. Understanding this pattern is crucial because many binary tree problems require operations that focus on these paths. For instance, tasks like finding the sum of all left leaves or determining specific paths that meet a given condition rely heavily on this approach. Mastering this pattern allows you to handle a wide variety of problems involving tree traversal and leaf node operations.

In this pattern, the general approach involves recursive traversal of the binary tree, where you explore each possible path from the root to the leaf nodes. During the traversal, you can perform specific operations, such as accumulating sums, validating path properties, or collecting nodes. The key is to ensure that you correctly identify and process each path leading to a leaf, as the operations often depend on the properties of these paths.

Let's explore the example problem and learn to solve root to leaf path pattern related problems.

Problem Statement

Given a binary tree where each node contains an integer, return the maximum sum obtained by following any path from the root node to a leaf node.

The path must start at the root and end at a leaf.

Examples

Example 1:

• Input: root = [8, 4, 9, 1, 6]

• Expected Output: 18
• Justification: The maximum sum is obtained by following the path 8 -> 4 -> 6, which sums to 18.

Example 2:

• Input: root = [5, 3, 7, null, null, 8, 2]

• Expected Output: 20
• Justification: The maximum sum is obtained by following the path 5 -> 7 -> 8, which sums to 20.

Solution

To solve root to leaf pathrelated problems, the most effective approach is to use a recursive depth-first search (DFS) traversal of the binary tree. In this case, the idea is to explore all possible paths from the root to the leaves while keeping track of the sum of values along each path. As we traverse, we accumulate the sum and compare it to the maximum sum found so far. This approach is efficient because it ensures that every possible path is considered, and the maximum path sum is captured by the time the traversal is complete.

Since we are only concerned with paths that start at the root and end at the leaves, a DFS traversal allows us to explore all potential paths systematically, ensuring no path is overlooked. This ensures an optimal solution with a time complexity proportional to the number of nodes in the tree.

Step-by-step Algorithm

1. Initialize Variables:

   • Create a variable maxSum and set it to the smallest possible integer value (Integer.MIN_VALUE in Java).
   • This variable will store the maximum path sum encountered during the traversal.

2. Define Recursive Function (findMaxSum):

   • The function takes two arguments: the current TreeNode and the currentSum (the sum of the values along the current path).

3. Base Case:

   • If the current TreeNode is null, return immediately. This handles cases where the tree branch ends.

4. Update Current Path Sum:

   • Add the value of the current node to currentSum.

5. Check for Leaf Node:

   • A node is a leaf if both its left and right children are null.
   • If the current node is a leaf, compare currentSum with maxSum:
     • If currentSum is greater than maxSum, update maxSum with the value of currentSum.

6. Recursive Calls:

   • If the current node has a left child:
     • Recursively call findMaxSum on the left child, passing the updated currentSum.
   • If the current node has a right child:
     • Recursively call findMaxSum on the right child, passing the updated currentSum.

7. Return Result:

   • After the recursion completes, return the value stored in maxSum as the result.

Algorithm Walkthrough

• Step 1:

  • Start at the root node with value 8.
  • Initialize maxSum to Integer.MIN_VALUE.
  • Call findMaxSum(root, 0) with root pointing to the node with value 8 and currentSum set to 0.

• Step 2:

  • At the node 8, add its value to currentSum: currentSum = 0 + 8 = 8.
  • Since node 8 is not a leaf, proceed to check its children.
  • Call findMaxSum on the left child, node 4, with currentSum = 8.

• Step 3:

  • At node 4, add its value to currentSum: currentSum = 8 + 4 = 12.
  • Since node 4 is not a leaf, proceed to check its children.
  • Call findMaxSum on the left child, node 1, with currentSum = 12.

• Step 4:

  • At node 1, add its value to currentSum: currentSum = 12 + 1 = 13.
  • Since node 1 is a leaf (both children are null), compare currentSum with maxSum.
  • maxSum is updated from Integer.MIN_VALUE to 13.
  • Return from this recursive call.

• Step 5:

  • Back at node 4, call findMaxSum on the right child, node 6, with currentSum = 12.

• Step 6:

  • At node 6, add its value to currentSum: currentSum = 12 + 6 = 18.
  • Since node 6 is a leaf, compare currentSum with maxSum.
  • maxSum is updated from 13 to 18.
  • Return from this recursive call.

• Step 7:

  • Back at node 8, call findMaxSum on the right child, node 9, with currentSum = 8.

• Step 8:

  • At node 9, add its value to currentSum: currentSum = 8 + 9 = 17.
  • Since node 9 is a leaf, compare currentSum with maxSum.
  • maxSum remains 18 because 17 is less than 18.
  • Return from this recursive call.

• Step 9:

  • The recursion completes, and the final maxSum is 18.
  • Return 18 as the result.

Code

// class TreeNode {
//     int val;
//     TreeNode left;
//     TreeNode right;

//     // Constructor to initialize a tree node with a value
//     TreeNode(int x) {
//         val = x;
//     }
// }

class Solution {

  private int maxSum;

  public int maxPathSum(TreeNode root) {
    // Initialize maxSum to the smallest possible value
    maxSum = Integer.MIN_VALUE;

    // Start the recursive process to find the maximum sum
    findMaxSum(root, 0);
    return maxSum;
  }

  private void findMaxSum(TreeNode node, int currentSum) {
    // Base case: if the current node is null, return immediately
    if (node == null) return;

    // Add the current node's value to the current path sum
    currentSum += node.val;

    // Check if the current node is a leaf
    if (node.left == null && node.right == null) {
      // Update maxSum if the current path sum is greater
      maxSum = Math.max(maxSum, currentSum);
    }

    // Recursively process the left and right subtrees
    findMaxSum(node.left, currentSum);
    findMaxSum(node.right, currentSum);
  }

  public static void main(String[] args) {
    // Create the binary tree for the first example
    TreeNode root = new TreeNode(8);
    root.left = new TreeNode(4);
    root.left.left = new TreeNode(1);
    root.left.right = new TreeNode(6);
    root.right = new TreeNode(9);

    Solution solution = new Solution();
    System.out.println(solution.maxPathSum(root)); // Output: 18
  }
}

Complexity Analysis

Time Complexity

The time complexity of the algorithm is , where N is the number of nodes in the binary tree.

• The algorithm uses a depth-first search (DFS) traversal to explore each node in the tree exactly once. In the worst case, every node from the root to the leaf must be visited, which results in linear time complexity relative to the number of nodes.

Space Complexity

The space complexity of the algorithm is , where H is the height of the binary tree.

• The space complexity is primarily determined by the recursive call stack. In the worst-case scenario (e.g., a skewed tree), the height of the tree could be N (when the tree is a straight line). However, for a balanced tree, the height is log(N). Thus, in the worst case, the space complexity is , but for a balanced tree, it is .

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