[LeetCode] 113. Path Sum II

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题目：

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]]

答案：

典型的DFS，可以用回溯的方式去解。

参考代码：

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

延伸思考：如果不要求一定是根到叶子的路径，而是任意两个节点的路径应该怎么解呢？

参考解法：（代码没有经过测试）

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> pathSum(TreeNode root, int sum) {        List<List<Integer>> ans = new ArrayList<>();        if (root != null) {            ArrayList<Integer> sums = new ArrayList<>();            sums.add(0); // 可以添加任意一个值            solve(ans, root, new ArrayList<>(), sums, sum);        }        return ans;    }    private void solve(List<List<Integer>> ans, TreeNode root, List<Integer> cur, List<Integer> sums, int target) {        if (root != null) {            cur.add(root.val);            int newSum = root.val + sums.get(sums.size()-1);            for (int i = 0; i < sums.size(); i++) {                if (newSum - sums.get(i) == target) {                    ans.add(new ArrayList<>(cur.subList(i, cur.size())));                }            }            sums.add(newSum);            solve(ans, root.left, cur, sums, target);            solve(ans, root.right, cur, sums, target);            sums.remove(sums.size() - 1);            cur.remove(cur.size() - 1);        }    }}

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