problem: 0437 path sum III

Author: cychiu8

0437 Path Sum III/0437 Path Sum III.md

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+# 0437. Path Sum III
+
+* Difficulty: medium
+* Link: https://leetcode.com/problems/path-sum-iii/
+* Topics: DFS-BFS
+* highlight: 遍歷順序
+
+# Clarification
+
+1. Check the inputs and outputs
+
+# Solution (DFS)
+
+### Thought Process
+
+- 每個點都當成一棵樹的 root
+- Implement
+    
+    ```python
+    
+    ```
+    
+
+### Complexity
+
+- Time complexity:
+- Space complexity:
+
+# Improvement (DFS + HashMap)
+
+### Thought Process
+
+- HashMap 紀錄前面路徑合出現過的次數
+    
+    ![Untitled](./Untitled.png)
+    
+- Implement
+    
+    ```python
+    # Definition for a binary tree node.
+    # class TreeNode:
+    #     def __init__(self, val=0, left=None, right=None):
+    #         self.val = val
+    #         self.left = left
+    #         self.right = right
+    class Solution:
+        count = 0
+        pathSumMap = {}
+        pathSumMap[0]=1
+        
+        def pathSum(self, root: Optional[TreeNode], targetSum: int) -> int:
+            
+            if not root:
+                return 0
+                   
+            def dfs(root, curSum):
+                if not root:
+                    return
+                curSum += root.val
+                
+                # 計算路路徑長度差
+                subPath = curSum - targetSum
+                self.count += self.pathSumMap.get(subPath, 0)
+                self.pathSumMap[curSum] = self.pathSumMap.get(curSum, 0) + 1
+    
+                dfs(root.left, curSum)
+                dfs(root.right, curSum)
+    
+                # 換另一條路下去前，要把先前的路刪掉
+                self.pathSumMap[curSum] = self.pathSumMap.get(curSum, 0) - 1
+                curSum -= root.val
+                
+            dfs(root, 0)
+            
+            return self.count
+    ```
+    
+
+### Complexity
+
+- Time complexity:
+- Space complexity:
+
+# Note