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.
Solution:
/**
* 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) {
return hasPathSum(root, sum, 0);
}
public boolean hasPathSum(TreeNode root, int sum, int tempSum)
{
if(root == null)
return false;
if( root.left==null && root.right == null)
return tempSum + root.val == sum;
return hasPathSum(root.left, sum, tempSum+root.val) || hasPathSum(root.right, sum, tempSum+root.val);
}
}
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