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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);
}
}
Java for LeetCode 113 Path Sum II
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原文地址：http://www.cnblogs.com/tonyluis/p/4524858.html
