title: Path Sum
tags:
- path-sum
- No.112
- simple
- tree
- recursive

Description

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.

Corner Cases

• empty tree

Solutions

Recursively reduce the value of the current node from sum. The running time is the same as BFS, O(V):

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {return false;}        
        return f(root, sum);
    }
    
    private boolean f(TreeNode x, int s) {
        if (x.left == null && x.right == null && x.val == s) {return true;}
        
        boolean fl = false;
        boolean fr = false;
        if (x.left  != null) {fl = f(x.left,  s - x.val);}
        if (x.right != null) {fr = f(x.right, s - x.val);}        
        return fl || fr;
    }
}