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]
]

问题描述：给出一个二叉树和整数sum，求和等于sum的所有路径列表。

问题解决：根据path sum问题，对每一个节点，除了维护一个nodes, vals LinkedList聊表外，再加一个路径列表，当和为sum时，将这个节点对应的路径加入结果列表中，当所有路径遍历完毕时，返回结果列表。

解法一：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Solution {public List<List<Integer>> pathSum(TreeNode root, int sum) {LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();LinkedList<Integer> vals = new LinkedList<Integer>();LinkedList<List<Integer>> lc = new LinkedList<List<Integer>>(); //每个节点保存的路径列表
          List<List<Integer>> l = new ArrayList<List<Integer>>();//保存最终结果List<Integer> temp = new ArrayList<Integer>(); if(root==null)return null;nodes.add(root);vals.add(root.val);temp.add(root.val);lc.add(temp);while(!nodes.isEmpty()){TreeNode cur = nodes.poll();int num = vals.poll();temp = lc.poll();if((cur.left==null && cur.right==null) && num==sum){l.add(temp);}if(cur.left!=null){nodes.add(cur.left);vals.add(num+cur.left.val);temp.add(cur.left.val);lc.add(temp);}if(cur.right!=null){nodes.add(cur.right);vals.add(num+cur.right.val);temp.add(cur.right.val);lc.add(temp);}}return l;}
}

但是这种方法空间。时间开销过大，需要一种新的方法改进。

解法二：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Solution {public List<List<Integer>> pathSum(TreeNode root, int sum) {List<List<Integer>> res = new ArrayList<List<Integer>>();//保存最终结果List<Integer> curli = new ArrayList<Integer>();int curVal = 0;getPath(root, sum, curVal, res, curli);return res;}public void getPath(TreeNode root, int sum, int curVal,List<List<Integer>> res, List<Integer> curli) {// TODO Auto-generated method stubif(root==null)return;curVal += root.val;curli.add(root.val);if((root.left==null && root.right==null) && curVal==sum){res.add(new ArrayList(curli));return;}if(root.left!=null){getPath(root.left, sum, curVal, res, curli);curli.remove(curli.size()-1);}if(root.right!=null){getPath(root.right, sum, curVal, res, curli);curli.remove(curli.size()-1);}}
}

转载于:https://www.cnblogs.com/little-YTMM/p/4531115.html

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