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.
Note: A leaf is a node with no children.
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.
题意:
给定二叉树和一个值,判断从根到叶是否存在一条路径,其路径和等于该值。
思路:
dfs
代码:
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null) return false; if(root.left== null && root.right== null)
return root.val == sum; if(hasPathSum(root.left, sum - root.val))
return true;
if(hasPathSum(root.right, sum - root.val))
return true; return false;
}
}