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

从任意一点dfs, 和dfs return 当前, 左节点, 右节点的模板

The basic idea is to subtract the value of current node from sum until  the subtraction equals 0, then we know that we got a path. and because the start node can be any node, so we should add the return value from its subtree. /**

 * 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;
        //helper(root,sum) 当前节点开始
        //pathSum(root.left,sum) 当前节点左节点开始
        //pathSum(root.right,sum) 当前节点右节点开始
        return helper(root,sum)+pathSum(root.left,sum)+pathSum(root.right,sum);
       
    }
    private int helper(TreeNode root,int sum){
       if(root==null) return 0;
       int count=0;
       if(root.val==sum) count++;
       return count+helper(root.left,sum-root.val)+helper(root.right,sum-root.val);
    }

}

要学会从任意节点遍历+ 分治法(后序遍历)

当前节点是否满足题意加上左子树和右子树的结果值
return count+helper(root.left,sum-root.val)+helper(root.right,sum-root.val);