Path Sum III
2017-11-16 10:36
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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.
题意:找出所以路径和为sum的路径总个数,不要求起点为根节点也不要求终点为叶子节点,只有一个要求:就是必须是从上往下遍历
解析:这题目标着easy,思想也确实简单,能想到用dfs遍历即可,那它既然不要求起点为根节点,就是说需要对每个节点都进行一次dfs。但是虽然知道这个思路,但是我却没能很好的写出来,看了solution才恍悟可以这么去写。
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.
题意:找出所以路径和为sum的路径总个数,不要求起点为根节点也不要求终点为叶子节点,只有一个要求:就是必须是从上往下遍历
解析:这题目标着easy,思想也确实简单,能想到用dfs遍历即可,那它既然不要求起点为根节点,就是说需要对每个节点都进行一次dfs。但是虽然知道这个思路,但是我却没能很好的写出来,看了solution才恍悟可以这么去写。
//pathSum表示对一个节点的DFS //solver表示dfs的递归式 public int pathSum(TreeNode root, int sum) { if (root == null) return 0; return soler(root,sum) + pathSum(root.left,sum) + pathSum(root.right,sum); } private int soler(TreeNode root,int sum) { if (root == null) return 0; return (root.val == sum? 1 : 0) + soler(root.left,sum - root.val) + soler(root.right,sum - root.val); }
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