0437. Path Sum III (M)

Path Sum III (M)

题目

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

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题意

在给定的树中找到一条从上到下的路径，使其和为给定值。路径的起点和终点可以是任意结点。

思路

双重递归。第一重递归确定根结点，限制需要查找的子树；第二重递归在该子树中找到从子树根结点出发和为sum的路径的个数。

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代码实现

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int pathSum(TreeNode root, int sum) {
        if (root == null) {
            return 0;
        }
        return findPath(root, 0, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }

    private int findPath(TreeNode root, int curSum, int sum) {
        if (root == null) {
            return 0;
        }
        curSum += root.val;
        return (curSum == sum ? 1 : 0) + findPath(root.left, curSum, sum) + findPath(root.right, curSum, sum);
    }
}