LeetCode(35)-Path Sum
題目：
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.
思路：
題意是要從根幾點root到葉子節點的所有組合，是不是等於給定的整數sum 考慮用遞歸，把對當前的判斷，轉化為root.left和sum -root.val的比較以及toot.right和sum-root.val的比較 -
代碼：
/**
* 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;
}
int count = 0;
if(root != null){
count = root.val;
}
if(root.left != null && root.right != null){
return hasPathSum(root.left,sum-count) || hasPathSum(root.right,sum-count);
}
if(root.right != null){
return hasPathSum(root.right,sum-count);
}
if(root.left != null){
return hasPathSum(root.left,sum-count);
}
if(count == sum){
return true;
}
return false;
}
}
