437-Path Sum III

Description

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.

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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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問題描述

給定二叉樹， 每個節點有一個整數值

找出路徑上的節點之和為給定值的路徑個數

路徑不一定以根節點為開始節點， 以葉子節點為結束節點， 但是必須從父節點到孩子節點(即從上往下)

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問題分析

分為兩個方法來做

find()找出以當前節點為開始節點的滿足條件的路徑的個數

pathSum()先序遍歷root, 用find()對每個節點計數並累加

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解法

class Solution {

    public int pathSum(TreeNode root, int sum) {
        if(root == null) return 0;

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


    public int find(TreeNode root, int sum){
        if(root == null) return 0;

        return (sum == root.val ? 1 : 0) + find(root.left, sum - root.val)  + find(root.right, sum - root.val);
    }
}