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112 Path Sum

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.

:::tip Thoughts This problem could be using a modified DFS. BFS would also work on this problem. :::

Solution: Recursion

Time complexity: O(n) Space complexity: O(n)

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        if (root.left == null && root.right == null) {
            return sum == root.val;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }
}

// Running time: 4 ms
class Solution {
public:
  bool hasPathSum(TreeNode* root, int sum) {
    if (!root) return false;
    if (!root->left && !root->right) return root->val==sum;
    int new_sum = sum - root->val;
    return hasPathSum(root->left, new_sum) || hasPathSum(root->right, new_sum);
  }
};
    1. Path Sum II
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