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.
- 题目大意：从给定的二叉树中找到root - to -leaf的path，并且path上面的数字之和等于给定的sum。
public class PathSum {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null)
return false;
//如果左右结点为null，此时到达叶子结点
if (root.left == null && root.right == null && sum - root.val == 0)
return true;
//递归判断
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
}
