112. Path Sum

112. Path Sum

 

https://www.youtube.com/watch?v=zdClzfnkvDY

 

 

The basic idea is to subtract the value of current node from sum until it reaches a leaf node and the subtraction equals 0, then we know that we got a hit. Otherwise the subtraction at the end could not be 0.

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
    
        if(root.left == null && root.right == null && sum - root.val == 0) return true;
    
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

 

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.