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LeetCode - 437. Path sum III

时间:2018-10-15 18:16:12      阅读:167      评论:0      收藏:0      [点我收藏+]

标签:contains   val   binary   bin   efi   int   bsp   ini   turn   

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

遍历整棵树,每个节点找一遍
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int cnt = 0;
    public int pathSum(TreeNode root, int sum) {
        if (root == null)
            return 0;
        find(root, sum);
        pathSum(root.left, sum);
        pathSum(root.right, sum);
        return cnt;
    }
    
    private void find(TreeNode node, int sum) {
        if (node == null)
            return;
        if (sum == node.val)
            cnt ++;
        find(node.left, sum-node.val);
        find(node.right, sum-node.val);
    }
}

 

LeetCode - 437. Path sum III

标签:contains   val   binary   bin   efi   int   bsp   ini   turn   

原文地址:https://www.cnblogs.com/wxisme/p/9792134.html

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