leetcode437. Path Sum III-优快云博客

本文链接：https://blog.youkuaiyun.com/rxt2012kc/article/details/72852257

本文介绍了一个二叉树中寻找特定路径的问题：给定一棵二叉树及一个整数目标值，在树中寻找所有从任意节点出发向下到任意其他节点的路径，使得这些路径上的节点值之和等于给定的目标值。文章详细阐述了解决这个问题的一种深度优先搜索（DFS）方法。

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437. Path Sum III

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

解法

dfs，最终结果为每一个结点都当做根节点，计算以该结点为根节点时满足条件的数量。

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

        return helper(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
    }

    public int helper(TreeNode root, int sum) {
        if (root == null) {
            return 0;
        }
        if (root.val == sum) {
            return 1 + helper(root.left, sum - root.val)
                    + helper(root.right, sum - root.val);
        }
        return helper(root.left, sum - root.val)
                + helper(root.right, sum - root.val);
    }
}