题目描述
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,
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
代码
static boolean hasPath;
public static boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
hasPath = false;
help(root, 0, sum);
return hasPath;
}
static void help(TreeNode node, int cur, int sum) {
cur += node.val;
boolean isLeaf = (node.left == null) && (node.right == null);
if (cur == sum && isLeaf) {
hasPath = true;
}
if (node.left != null) {
help(node.left, cur, sum);
}
if (node.right != null) {
help(node.right, cur, sum);
}
cur -= node.val;
}
更简洁的代码:
public static boolean hasPathSum2(TreeNode root, int sum) {
if (root == null) {
return false;
}
if (root.left == null && root.right == null) {
return root.val == sum;
}
return (root.left != null && hasPathSum2(root.left, sum - root.val))
|| (root.right != null && hasPathSum2(root.right, sum
- root.val));
}