看了一下数据结构中树的操作，A这题感觉好一点了。Symmetric Tree和这个很相似来着，可以借鉴一下这个思路。用递归处理
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.
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * 7 * TreeNode right; 8 * TreeNode(int x) { val = x; } 9 * } 10 */ 11 public class Solution { 12 boolean flag = false; 13 14 public boolean hasPathSum(TreeNode root, int sum) { 15 if(null == root) 16 return false; 17 hasPathSum1(root, sum); 18 return flag; 19 } 20 public boolean hasPathSum1(TreeNode root, int sum) { 21 if(null == root.left && null == root.right){ 22 if(sum == root.val) 23 flag = true; 24 return sum == root.val; 25 } 26 27 else 28 return hasPathSum(root.left, sum - root.val) || 29 30 hasPathSum(root.right, sum - root.val); 31 } 32 }
PS:这样也是可以的
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * 7 * TreeNode right; 8 * TreeNode(int x) { val = x; } 9 * } 10 */ 11 public class Solution { 12 public boolean hasPathSum(TreeNode root, int sum) { 13 if(null == root) 14 return false; 15 if(null == root.left && null == root.right){ 16 17 return sum == root.val; 18 } 19 20 else 21 return hasPathSum(root.left, sum - root.val) || 22 23 hasPathSum(root.right, sum - root.val); 24 } 25 }