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.
一刷
题解：这题最容易出错的部分是要求从root到leaf存在一个path, 从而判断是否是leaf的方式：root.left == null && root.right == null && sum == root.val
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public 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);
}
}
二刷
思路同上
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public 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);
}
}
三刷
recursion
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if(root == null) return false;
if(root.val == sum && root.left == null && root.right == null) return true;
return hasPathSum(root.left, sum-root.val) ||
hasPathSum(root.right, sum-root.val);
}
}
