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;		}	}