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LeetCode 437. Path Sum III

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Question

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

Explanation:

这道题给了二叉树和一个sum，找出所有子路径和为sum的路径。 直接用递归方法做，把每个点都当做父节点遍历，找到存在的路径，相加即可。 相关题目：

  • Leetcode 64. Minimum Path Sum
  • Leetcode 112. Path Sum
  • Leetcode 113. Path Sum II

之后再对这类题目进行总结。

Code: Java 

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void flatten(TreeNode root) {
        if (root == null) return;
        TreeNode right = root.right;
        TreeNode left = root.left;
        flatten(right);
        flatten(left);
        root.left = null;
        root.right = left;
        TreeNode cur = root;
        while(cur.right != null) cur = cur.right;
        cur.right = right;
    }
}

Python 

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class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: int
        """
        if not root:
            return 0
        return self.pathHelper(root, sum) + self.pathSum(root.left, sum) + self.pathSum(root.right, sum)
    def pathHelper(self, root, sum):
        result = 0
        if not root:
            return 0
        if (sum == root.val):
            result += 1
        result += self.pathHelper(root.left, sum - root.val)
        result += self.pathHelper(root.right, sum - root.val)
        return result