437. Path Sum III
阿新 • • 發佈:2017-10-21
star nta int ont pub binary ber from ins
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
題目含義:給定一棵二叉樹,以及一個和sum,問在樹中是否存在一條路徑,路徑上所有結點之和恰好等於sum。路徑不限制一定要開始於根結點或者結束於葉子結點,但是一定要是向下的(從父親結點向兒子結點)。如果存在這樣的路徑,求出有多少條
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 publicint pathSum(TreeNode root, int sum) { 9 // 以每一個節點作為路徑根節點進行前序遍歷,查找每一條路徑的權值和與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