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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.
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] ]
具体思路和代码如下:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
/**
* 回溯法求解
* 整体思想是遍历,然后添加list逐一试探
* 符合要求的添加结果集
* 不符合要求的删除,然后回溯
*/
List<List<Integer>> list = new ArrayList<List<Integer>>();
public List<List<Integer>> pathSum(TreeNode root, int sum) {
if(root == null){
return list;
}
path(root,sum,new ArrayList<Integer>());
return list;
}
private void path(TreeNode root,int sum, List<Integer> al){
if(root == null){
return;
}
if(root.val == sum){
if(root.left == null && root.right == null){
al.add(root.val);
//添加结果一定要重新生成实例
list.add(new ArrayList<Integer>(al));
al.remove(al.size()-1);//删除
return;
}
}
al.add(root.val);
path(root.left,sum - root.val,al);
path(root.right,sum - root.val,al);
al.remove(al.size()-1);//一定要删除,确保回溯准确
}
}