[Leetcode]Path Sum&Path Sum II
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Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:Given the below binary tree and
sum = 22
,5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
最简单的递归题
public boolean hasPathSum(TreeNode root, int sum) { if (root == null) return false; if (root.left == null && root.right == null && root.val == sum) { return true; } return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val); }
Path Sum II
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:Given the below binary tree and
sum = 22
,5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5]]
两道题思路是一样一样的,都是利用dfs自顶向下搜索。不同的事这里需要维护一个结果集,因此需要在每次递归中加入正在递归的节点,如果该节点的值正好等于我们需要的sum并且是叶子结点,那说明找到了一种方案,将此方案加入结果集中。如果还有左右子节点,则递归遍历这些节点,如不符合即返回并从结果中删去这个遍历过的节点。
public List<List<Integer>> pathSum(TreeNode root, int sum) { List<List<Integer>> res = new ArrayList<List<Integer>>(); if (root == null) { return res; } List<Integer> item = new ArrayList<Integer>(); // item.add(root.val); helper(root, res, item, sum); return res; } private void helper(TreeNode root, List<List<Integer>> res, List<Integer> item, int sum) { if (root == null) { return; } item.add(root.val); if (root.left == null && root.right == null && sum == root.val) { res.add(new ArrayList<Integer>(item)); return; } if (root.left != null) { helper(root.left, res, item, sum - root.val); item.remove(item.size() - 1); } if (root.right != null) { helper(root.right, res, item, sum - root.val); item.remove(item.size() - 1); } }
这里值得一提的是这个问题和 Triangle问题有点相似,都是求自顶向下的路径和。但是triangle问题给的是结构化的一个三角形,并且求的是最小路径和,所以用dp填表法来做。
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