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.
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public bool HasPathSum(TreeNode root, int sum) {
if(root == null){
return false;
}
if(root.left == null && root.right == null){
return root.val==sum;
}
return HasPathSum(root.left, sum-root.val)||HasPathSum(root.right,sum-root.val);
}
}Time Complexity: O(n)
Space Complexity: O(1)
