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]
]
 

解题思路:

DFS,JAVA实现如下:

	static public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> list = new ArrayList<List<Integer>>();
List<Integer> alist = new ArrayList<Integer>();
if (root != null)
dfs(alist,list, root, sum);
return list;
} public static void dfs(List<Integer> alist,List<List<Integer>> list, TreeNode root, int sum) {
if (root.left == null && root.right == null) {
if (sum == root.val) {
alist.add(root.val);
list.add(new ArrayList<Integer>(alist));
alist.remove(alist.size() - 1);
}
return;
}
alist.add(root.val);
if (root.left == null)
dfs(alist,list, root.right, sum - root.val);
else if (root.right == null)
dfs(alist,list, root.left, sum - root.val);
else {
List<Integer> alistCopy = new ArrayList<Integer>(alist);
dfs(alist,list, root.right, sum - root.val);
alist=alistCopy;
dfs(alist,list, root.left, sum - root.val);
} }

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