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Path Sum
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 }
Path Sum
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