leetcode 437 路径和 Path Sum III

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int pathSum(TreeNode* root, int sum) {
13         queue<TreeNode*> q;
14         int dp[1000];
15         q.push(root);
16         int index=0;
17         int count=0;
18         while(!q.empty()){
19             TreeNode* temp=q.front();
20             q.pop();
21             q.push(temp->left);
22             q.push(temp->right);
23             if(temp==NULL) dp[index++]=1000010;
24             else dp[index++]=temp->val;
25         }
26         for(int i=0;i<index;i++){
27             if(i==0&&dp[i]==sum){
28                 count++;
29             }else if(dp[i]<=1000000&&dp[i]>=-1000000){
30                 dp[i]=dp[(i-1)/2]+dp[i];
31                 if(dp[i]==sum) count++;
32             }
33         }
34         return count;
35     }
36 };

想使用层序遍历+动态规划的方法O(n)完成，被NULL节点不能加入queue<TreeNode*>  q给卡住了，之后再看看怎么改；对这种含有NULL多的怎么层序啊啊啊啊啊啊；

先看看大佬的方法：

python：非递归先序遍历+字典

 1 # Definition for a binary tree node.
 2 # class TreeNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7
 8 class Solution(object):
 9     def pathSum(self,root,sum):
10         if root==None:
11             return 0
12         from collections import defaultdict
13         dictionary =defaultdict(int)
14         dictionary[0]=1
15         pNode,curr_sum,stack,res,prev=root,0,[],0,None
16         while(len(stack) or pNode):
17             if pNode:
18                 curr_sum+=pNode.val
19                 stack.append([pNode,curr_sum])
20                 dictionary[curr_sum]+=1
21                 pNode=pNode.left
22             else:
23                 pNode,curr_sum=stack[-1]
24                 if pNode.right==None or pNode.right==prev:
25                     res+=dictionary[curr_sum-sum]
26                     if sum==0:
27                         res-=1
28                     dictionary[curr_sum]-=1
29                     stack.pop()
30                     prev=pNode
31                     pNode=None
32                 else:
33                     pNode=pNode.right
34         return res
35     

原文地址：https://www.cnblogs.com/joelwang/p/10453023.html