LeetCode——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]
]

给定一个二叉树和一个值。找出全部根到叶的路径和等于这个值的路径。

深度优先遍历。

	public List<List<Integer>> pathSum(TreeNode root, int sum) {
		List<List<Integer>> ret = new ArrayList<List<Integer>>();
		List<Integer> list = new ArrayList<Integer>();
		dfs(root,sum,ret,list);
		return ret;
	}
	public void dfs(TreeNode root,int sum,List<List<Integer>> ret,List<Integer> list){
		if(root == null)
			return ;
		if(root.val == sum && root.left == null && root.right == null){
			list.add(root.val);
			List<Integer> temp = new ArrayList<Integer>(list);//拷贝一份
			ret.add(temp);
			list.remove(list.size() - 1);//再删除
			return ;
		}
		list.add(root.val);
		dfs(root.left,sum-root.val,ret,list);
		dfs(root.right,sum-root.val,ret,list);
		list.remove(list.size() - 1);
	}
	// Definition for binary tree
	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}