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.
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.
中文理解:给定一个二叉树,判断这个二叉树是否存在一条从根到叶节点的节点值之和为target值的路径,若存在返回true,否则返回false。
解题思路:采用递归思想,若根节点为null,返回false;若根节点为叶子节点,节点值为target返回true,否则返回false;若根节点不为叶子节点,则递归hasPath(root.left,target-root.val) || hasPath(root.right,target-root.val)。
代码(java):
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root==null)return false;
else if(root.right==null && root.left==null){
if(root.val==sum)return true;
else return false;
}
else{
return hasPathSum(root.left,sum-root.val) || hasPathSum(root.right,sum-root.val);
}
}
}