113. Path Sum II

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]

]

链接： http://leetcode.com/problems/path-sum-ii/

题解：

与上一题不同的是，要存储所有的root - leaf路径，所以在DFS的基础上我们还要增加backtracking。回溯的点有两个，一个是当前root满足条件，返回时，此时回溯保证当前root的sibling结果正确；另外一个是遍历完当前root的左子树和右子树时，这时回溯也是保证sibling结果正确。比如 root = [0, 1, 1], sum = 1，则计算完左边的1时可以正确计算到右边的1。或者 root = [0, 1, 1, 0, 0]， sum = 1，遍历完左边的两个0后可以正确计算到右边1的结果。 (有空要组织一下语言)

Time Complexity - O(n)， Space Complexity - O(n)。

/**

 * 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>> res = new ArrayList<>();

        ArrayList<Integer> list = new ArrayList<>();

        dfs(res, list, root, sum);

        return res;

    }

    private void dfs(List<List<Integer>> res, ArrayList<Integer> list, TreeNode root, int sum) {

        if(root == null)

            return;

        if(root.left == null && root.right == null && root.val == sum) {

            list.add(root.val);

            res.add(new ArrayList<Integer>(list));

            list.remove(list.size() - 1);

            return;

        }

        list.add(root.val);

        sum -= root.val;

        dfs(res, list, root.left, sum);

        dfs(res, list, root.right, sum);

        list.remove(list.size() - 1);

    }

}

二刷:

注意回溯的点。

Java:

/**

 * 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>> res = new ArrayList<>();

        List<Integer> list = new ArrayList<>();

        pathSum(res, list, root, sum);

        return res;

    }

    private void pathSum(List<List<Integer>> res, List<Integer> list, TreeNode root, int sum) {

        if (root == null) return;

        sum -= root.val;

        list.add(root.val);

        if (root.left == null && root.right == null && sum == 0) {

            res.add(new ArrayList<>(list));

            list.remove(list.size() - 1);

            return;

        }

        pathSum(res, list, root.left, sum);

        pathSum(res, list, root.right, sum);

        list.remove(list.size() - 1);

    }

}

测试：

巴特西

113. Path Sum II

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