437. Path Sum III-优快云博客

本文介绍了一种解决二叉树中寻找特定和路径的方法。通过深度优先搜索(DFS)遍历二叉树的所有路径，并统计路径和等于给定值的情况。文章提供了详细的算法实现步骤及示例。

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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

没别的想法，就暴力dfs

注意节点的值可能有负数，所以不能随意剪枝

/**
 * 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;
		
		inorder(root, sum);
		
		return count;
	}
	
	int count=0;
	
	public void dfs(TreeNode root,int sum,int sumNow)
	{
		sumNow+=root.val;
		if(sumNow==sum)
			count++;
		if(root.left!=null)
			dfs(root.left, sum, sumNow);
		if(root.right!=null)
			dfs(root.right, sum, sumNow);
	}
	
	public void inorder(TreeNode root,int sum)
	{
		if(root.left!=null)
			inorder(root.left, sum);
		dfs(root, sum, 0);
		if(root.right!=null)
			inorder(root.right, sum);
		
	}
}