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  • DFS回溯只在递归基回溯————leetcode112

    # Definition for a binary tree node.
    # class TreeNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    # class Solution:
    #     def hasPathSum(self, root, sum):
    #         """
    #         :type root: TreeNode
    #         :type sum: int
    #         :rtype: bool
    #         """
    class Solution:
        
        res = False
        
        def hasPathSum(self, root, sum):
            """
            :type root: TreeNode
            :type sum: int
            :rtype: bool
            """
            if root is None:
                return False
            self.dfs2(root,0,sum)
            return self.res
    
        
    
        def dfs2(self, node, temp_sum, sum):
            temp_sum += node.val
            if not node.left and not node.right:
                if temp_sum == sum:
                    self.res = True
                    return
                temp_sum -= node.val
                return
            if node.left:
                self.dfs2(node.left,temp_sum,sum)
            if node.right:
                self.dfs2(node.right,temp_sum,sum)
            
                
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  • 原文地址:https://www.cnblogs.com/vector11248/p/9712554.html
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