LeetCode 437. Path Sum III (路径之和之三)
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10 / \ 5 -3 / \ \ 3 2 11 / \ \ 3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3 2. 5 -> 2 -> 1 3. -3 -> 11
题目标签:Tree
这道题目给了我们一个二叉树,和一个sum,让我们找到有多少条path 的和是等于sum的,这里的path不一定要从root 到底,可以是中间的点开始到下面的点。需要另外设两个functions。
traverseTree function - preOrder 遍历每一个点,然后把每一个点代入到findPathSum function。
findPathSum function - 找到从root 开始到各个点的 path sum, 和sum比较,一样的话,就res++。
利用traverseTree 来把每一个点(不同的level)代入findPathSum, 来找到从那个点开始往下的各种path sum,这样就可以包含 中间的点开始到下面的点的这种可能性。而且不会重复,因为每次代入的点,都是唯一的,而且找的path sum都是从这个点起始的。这两个funciton 就相当于 for loop 中间 再包括一个for loop, 来遍历array。
Java Solution:
Runtime beats 68.27%
完成日期:07/06/2017
关键词:Tree
关键点:利用两个递归functions来搜索从每一层level 各点开始到下面层次点的path值
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution 11 { 12 int res = 0; 13 public int pathSum(TreeNode root, int sum) 14 { 15 if(root == null) 16 return res; 17 18 traverseTree(root,sum); 19 20 return res; 21 } 22 23 public void traverseTree(TreeNode node, int sum) 24 { 25 if(node == null) 26 return; 27 28 findPathSum(node, 0, sum); 29 traverseTree(node.left, sum); 30 traverseTree(node.right, sum); 31 32 } 33 34 public void findPathSum(TreeNode node, int path_sum, int sum) 35 { 36 if(node == null) 37 return; 38 39 if(path_sum + node.val == sum) 40 res++; 41 42 findPathSum(node.left, path_sum + node.val, sum); 43 findPathSum(node.right, path_sum + node.val, sum); 44 45 } 46 }
参考资料:N/A
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