path-sum 3

https://leetcode.com/problems/path-sum-iii/submissions/

 

题目：

 

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int count = 0;
    public int pathSum(TreeNode root, int sum) {
　　　　　//第2个integer是该值的频次
        Map<Integer,Integer> map = new HashMap<>();
        //如果从root加到leaf正好= sum 就会命中
        map.put(0,1);
        findAllNodeSum(root,0,sum,map);
        return count;
    }
    
    void findAllNodeSum(TreeNode root,int cursum,int target,Map<Integer,Integer> map){
        if(root==null){
            return ;
        }
        cursum = cursum +root.val;
        if(map.containsKey(cursum-target)){
            count += map.get(cursum-target);
        }
        //判断是否以前节点的加现在的节点是否有符合要求的
        if(!map.containsKey(cursum)){
            map.put(cursum,1);
        }else{
            map.put(cursum,map.get(cursum)+1);
        }
        
        //dfs
        findAllNodeSum(root.left,cursum,target,map);
        findAllNodeSum(root.right,cursum,target,map);
        //把当前叶子节点的和清掉
        map.put(cursum,map.get(cursum)-1);
    }
    
    
}

如下图涉及自身为target和从root-leaf等于target2种情况

从root到leaf相加等于target被统计到count中是因为0，1在map中

 

这种情况是当前节点与上面的局部节点相加等于target，这种情况被统计到count，是因为5当时在map中