112. Path Sum (Easy) Microsoft

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 andsum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

Solution 1: DFS O(n); O(h)

    /**
     * Solution1: DFS O(n); O(h)
     * Subtract the value of current node from sum until it reaches a leaf node and sum equals 0.
     */
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }

        sum -= root.val; //先减然后和0比较。
        if (root.left == null && root.right == null && sum == 0) {
            return true;
        }

        return hasPathSum(root.left, sum) || hasPathSum(root.right, sum);
    }

另一种写法

    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
            return false;
        }
        if (root.left == null && root.right == null && root.val == sum) {
            return true;
        }
        return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
    }