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.
For 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.
HashMap solution implemted in Java
Pre-order DFS tree traversal could be very strait-forward for solving this problem.
Time complexity: O(n)
RUntime: 1ms
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
/**
* Returns boolean result whether any root-to-leaf path sums up to the "sum" given.
* @param TreeNode root of TreeNode type
* @param int int sum
* @return boolean result
*/
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) return false;
if ((sum-root.val == 0) && (root.left == null) && (root.right == null)) {
return true;
}
return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
}
}
Runtime: 1ms