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112.path sum

admin 11月 12, 2020 0

Question

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,

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7
 
      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.

Answer

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

Better Answer

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

Time complexity: $O(n)$

Space complexity: $O(n)$

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