Path Sum

##题目

####Path Sum

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.

##解题思路
这道题是树操作的题目,判断是否从根到叶子的路径和跟给定sum相同的。还是用常规的递归方法来做,递归条件是看左子树或者右子树有没有满足条件的路径,也就是子树路径和等于当前sum减去当前节点的值。结束条件是如果当前节点是空的,则返回false,如果是叶子,那么如果剩余的sum等于当前叶子的值,则找到满足条件的路径,返回true。

##算法代码
代码采用JAVA实现:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/

public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
//树为空
if(root ==null)
return false;
//已到叶子节点,且满足根结点的值等于sum
if(root.left==null && root.right==null && root.val==sum)
return true;
//判断左右子树是否有满足sum-root.val值的情况。
return hasPathSum(root.left,sum-root.val) || hasPathSum(root.right,sum-root.val);
}
}

Comments