437. Path Sum III

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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
题目含义：给定一棵二叉树，以及一个和sum，问在树中是否存在一条路径，路径上所有结点之和恰好等于sum。路径不限制一定要开始于根结点或者结束于叶子结点，但是一定要是向下的（从父亲结点向儿子结点）。如果存在这样的路径，求出有多少条

 1     int findPath(TreeNode node, int curSum, int sum) {
 2         if (node == null) return 0;
 3         curSum += node.val;
 4         int sameCount = curSum == sum ? 1 : 0;
 5         return sameCount + findPath(node.left, curSum, sum) + findPath(node.right, curSum, sum);
 6     }
 7     
 8     public int pathSum(TreeNode root, int sum) {
 9 //        以每一个节点作为路径根节点进行前序遍历，查找每一条路径的权值和与sum是否相等
10         if (root == null) return 0;
11         int res = findPath(root, 0, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);
12         return res;        
13     }

437. Path Sum III

标签：二叉树   blog   nod   contain   存在   parent   only   lin   pre   

原文地址：http://www.cnblogs.com/wzj4858/p/7705256.html