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;
}
}
