LeetCode112. Path Sum (思路及python解法)

本文探讨了在给定的二叉树中寻找一条从根节点到叶节点的路径,使得路径上所有节点的值之和等于给定的总和。通过三种方法:迭代、递归和深度优先搜索(DFS),详细解释了如何解决这个问题。

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Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

 

<思路>让child_value += father_val,一直到叶子节点,得到的就是整个路径上的和。看着很麻烦。

# 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 hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if root is None:
            return False
        
        value = []
        level = [root]
        
        while level:
            queue = []
            for node in level:
                if node.left:
                    node.left.val += node.val
                    queue.append(node.left)
                if node.right:
                    node.right.val += node.val
                    queue.append(node.right)
                if node.left is None and node.right is None:
                    value.append(node.val)
            level = queue
        
        if sum in value:
            return True 
        return False

看到discuss里一个递归的方法很精彩,学习借鉴。

# 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):
        if not root:
            return False

        if not root.left and not root.right and root.val == sum:
            return True
        
        sum -= root.val

        return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)

更新用DFS算法做的。 

class Solution:
    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
        if root is None:return 0
        stack=[(root, root.val)]
        
        while stack:
            root, rootval = stack.pop(0)
            if root.left is None and root.right is None:
                if rootval==sum:return True
                else: continue
            if root.left:
                stack.append((root.left, rootval+root.left.val))
            if root.right:
                stack.append((root.right, rootval+root.right.val))
        return False

 

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