What if you could instantly know if a path in a complex tree adds up to your target number without checking every branch yourself?
Why Path Sum Root to Leaf in Binary Tree in DSA Javascript?
Imagine you have a family tree drawn on paper, and you want to find if there is a path from the oldest ancestor to a youngest family member where the ages add up to a certain number.
Doing this by checking every possible path manually is like tracing every branch with a pencil and adding numbers each time.
Manually adding numbers along every path is slow and easy to make mistakes.
As the tree grows bigger, it becomes impossible to keep track of all paths and sums without missing or repeating steps.
Using a program to check paths in the tree automatically adds the numbers as it moves down each branch.
This way, it quickly finds if any path matches the target sum without checking paths again and again.
function checkPathSum(tree, target) {
// Manually trace each path and add values
// Very long and complex for big trees
}function hasPathSum(node, targetSum) {
if (!node) return false;
if (!node.left && !node.right) return node.val === targetSum;
return hasPathSum(node.left, targetSum - node.val) || hasPathSum(node.right, targetSum - node.val);
}This concept lets you quickly find if any root-to-leaf path in a tree adds up to a target number, enabling fast decision-making on complex tree data.
In a navigation app, finding a route where the total distance matches a user's preferred travel distance is like finding a path sum in a tree of roads.
Manual checking of all paths is slow and error-prone.
Path sum checking automates adding values along tree paths.
This helps quickly find if a path meets a target sum.
Practice
Solution
Step 1: Understand the problem statement
The problem asks if there is any path from the root node down to any leaf node where the sum of the node values equals a given target sum.Step 2: Differentiate from other tree properties
This is different from finding depth, counting nodes, or summing all nodes. It specifically checks for a path sum from root to leaf.Final Answer:
If there exists a root-to-leaf path with a sum equal to a given target -> Option AQuick Check:
Path sum problem = Check root-to-leaf sum [OK]
- Confusing path sum with total tree sum
- Checking any path, not root-to-leaf
- Ignoring leaf node condition
Solution
Step 1: Identify base case for recursion
When the current node is null, it means no path exists here, so return false to indicate failure to find the path.Step 2: Check JavaScript strict equality and return value
Using 'node === null' is correct to check null nodes, and returning false means no valid path found.Final Answer:
if (node === null) return false; -> Option DQuick Check:
Null node means no path = false [OK]
- Returning true on null node
- Using == instead of ===
- Returning null instead of boolean
class TreeNode {
constructor(val, left = null, right = null) {
this.val = val;
this.left = left;
this.right = right;
}
}
function hasPathSum(node, sum) {
if (node === null) return false;
if (!node.left && !node.right) return sum === node.val;
return hasPathSum(node.left, sum - node.val) || hasPathSum(node.right, sum - node.val);
}
const root = new TreeNode(5, new TreeNode(4, new TreeNode(11, new TreeNode(7), new TreeNode(2))), new TreeNode(8, null, new TreeNode(4, null, new TreeNode(1))));
console.log(hasPathSum(root, 22));Solution
Step 1: Trace the path sums in the tree
Check if any root-to-leaf path sums to 22. The path 5->4->11->2 sums to 22 (5+4+11+2=22).Step 2: Evaluate the recursive calls
The function returns true because the path exists, so console.log prints true.Final Answer:
true -> Option AQuick Check:
Path 5-4-11-2 sum=22 = true [OK]
- Ignoring leaf node condition
- Miscomputing sum along path
- Assuming false when path exists
function hasPathSum(node, sum) {
if (node === null) return false;
if (!node.left && !node.right) return sum = node.val;
return hasPathSum(node.left, sum - node.val) || hasPathSum(node.right, sum - node.val);
}Solution
Step 1: Identify incorrect assignment in leaf check
The line 'return sum = node.val;' uses assignment '=' instead of comparison '===' which causes wrong behavior.Step 2: Correct the comparison operator
Replace '=' with '===' to properly check if sum equals node.val at leaf nodes.Final Answer:
Change 'sum = node.val' to 'sum === node.val' in the leaf check -> Option CQuick Check:
Use '===' for comparison, not '=' [OK]
- Using '=' instead of '===' in conditions
- Removing necessary base cases
- Changing logical operators incorrectly
function hasPathSum(node, sum) {
if (node === null) return false;
if (!node.left && !node.right) return sum === node.val;
return hasPathSum(node.left, sum - node.val) || hasPathSum(node.right, sum - node.val);
}Solution
Step 1: Understand negative values impact
The current recursive approach subtracts node values from sum and checks leaf equality, which works even if values are negative.Step 2: Confirm no cycles in binary tree
Binary trees have no cycles, so no infinite loops occur; no visited set needed.Final Answer:
No change needed; current code works with negative values -> Option BQuick Check:
Negative values handled by subtraction correctly [OK]
- Adding unnecessary visited sets
- Changing base case incorrectly
- Using absolute values which breaks logic
Ready to prove it?
Pass the test and this pattern moves unknown
