LeetCode 113. Path Sum II（路径和）

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原题网址：https://leetcode.com/problems/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]]

方法：深度优先搜索。

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    private List<List<Integer>> paths = new ArrayList<>();    private void find(TreeNode root, List<Integer> values, int currentSum, int targetSum) {        if (root == null) return;        values.add(root.val);        currentSum += root.val;        if (root.left != null && root.right != null) {            find(root.left, values, currentSum, targetSum);            find(root.right, values, currentSum, targetSum);        } else if (root.left != null) {            find(root.left, values, currentSum, targetSum);        } else if (root.right != null) {            find(root.right, values, currentSum, targetSum);        } else if (currentSum == targetSum) {            List<Integer> path = new ArrayList<>(values.size());            path.addAll(values);            paths.add(path);        }        values.remove(values.size()-1);    }    public List<List<Integer>> pathSum(TreeNode root, int sum) {        find(root, new ArrayList<>(), 0, sum);        return paths;    }}

另一种实现：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    private List<List<Integer>> results = new ArrayList<>();    private void find(List<Integer> path, TreeNode root, int sum) {        if (root == null) return;        if (root.left == null && root.right == null && root.val == sum) {            List<Integer> result = new ArrayList<>();            result.addAll(path);            result.add(root.val);            results.add(result);            return;        }        path.add(root.val);        if (root.left != null) {            find(path, root.left, sum-root.val);        }        if (root.right != null) {            find(path, root.right, sum-root.val);        }        path.remove(path.size()-1);    }    public List<List<Integer>> pathSum(TreeNode root, int sum) {        find(new ArrayList<>(), root, sum);        return results;    }}

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