Path Sum II

Path Sum II

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Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.
Note: A leaf is a node with no children.

Example

input:[5,4,8,11,null,13,4,7,2,null,null,5,1] 22
output:[[5,4,11,2],[5,8,4,5]]

Solution

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @param sum, an integer
    # @return a list of lists of integers
    def pathSum(self, root, sum):
        def dfs(root, currsum, valuelist):
            if root.left==None and root.right==None:
                if currsum==sum: res.append(valuelist)
            if root.left:
                dfs(root.left, currsum+root.left.val, valuelist+[root.left.val])
            if root.right:
                dfs(root.right, currsum+root.right.val, valuelist+[root.right.val])
        
        res=[]
        if root==None: return []
        dfs(root, root.val, [root.val])
        return res

遇到一个很奇怪的问题，我用上一篇博客同样的方法去dfs，list列表居然不回溯，贴在这里，以后来看

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        self.num_list = []
        self.generate(root, [], sum, 0)
        return self.num_list
        
    def generate(self, root, temp, sum, count):
        if not root:
            return
        temp.append(root.val)
        if not root.left and not root.right:
            self.num_list.append(temp)
            return
        if root.left:
            self.generate(root.left, temp, sum, count+root.val)
        if root.right:
            self.generate(root.right, temp, sum, count+root.val)