LeetCode——Path Sum II

標籤：leetcode

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]]

給定一個二叉樹和一個值，找出所有根到葉的路徑和等於這個值的路徑。

深度優先遍曆。

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);//拷貝一份ret.add(temp);list.remove(list.size() - 1);//再刪除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 treepublic class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}}