Problem
1. 题目简述
给出一棵二叉树和一个sum值,根节点为root,求从root节点到叶子节点是否存在一条路径其节点之和为sum。例如:
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.
2. 算法思路
DFS
很简单的DFS的思路,中序遍历的思路,记录从根节点到当前节点的和,到达根节点时做一次判断。
3. 解法
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
if (root.left == null && root.right == null && sum == root.val) {
return true;
}
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}