Path Sum
Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.
A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.
Example 2:
Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.
Example 3:
Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.
Constraints:
The number of nodes in the tree is in the range [0, 5000].
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
思路一：递归统计每条路径的值
public boolean hasPathSum(TreeNode root, int targetSum) {
return hasPathSum(root, 0, targetSum);
}
public boolean hasPathSum(TreeNode root, int val, int target) {
boolean success = false;
if (root.left != null) {
success = hasPathSum(root.left, val + root.val, target);
}
if (root.right != null) {
success = success || hasPathSum(root.right, val + root.val, target);
}
if (root.left == null && root.right == null) {
return target == val + root.val;
}
return false;
}