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.
代碼:
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
class Solution {
/**
***採用遞歸的方法,我們用遞歸來實現!!!
*非常的簡單 :
*其基本思想是從總和中減去當前節點的值,
*直到它到達一個葉節點,結果等於0,那麼我們知道我們得到了一個命中。
*否則,最後的結果不能爲0。**
*/
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null) return false;
if(root.left == null && root.right == null && sum - root.val == 0) return true;
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}