Python实现"路径总和"的三种方法

最新推荐文章于 2024-04-13 21:14:27 发布

原创最新推荐文章于 2024-04-13 21:14:27 发布 · 2.1k 阅读

2 ·

CC 4.0 BY-SA版权

Algorithms 专栏收录该内容

106 篇文章

订阅专栏

本文探讨了如何判断一棵二叉树中是否存在从根节点到叶子节点的路径，其节点值之和等于给定的数。提供了三种方法：递归、求所有路径和及迭代，并附带Python实现。

给定一棵二叉树和一个总数，判断树中是否存在从根结点到叶子节点的累加值等于该给定的总数

注意：叶子节点没有子树

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.

1：递归，从最左边的路径开始，一旦满足要求就返回True

class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root:
            return False
        return self.recursionEachPath(root, 0, sum)
        
    def recursionEachPath(self, node, curSum, sum):
        if not node:
            return curSum==sum
        if not node.left and node.right is not None:
            return self.recursionEachPath(node.right, curSum+node.val, sum)
        if node.left is not None and not node.right:
            return self.recursionEachPath(node.left, curSum+node.val, sum)
        left = self.recursionEachPath(node.left, curSum+node.val, sum)
        right = self.recursionEachPath(node.right, curSum+node.val, sum)
        return left or right

该方法优化之后的代码为(参考他人代码)：

def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root:
            return False
        if not root.left and not root.right:
            return root.val==sum
        if not root.right:
            return self.hasPathSum(root.left, sum-root.val)
        if not root.left:
            return self.hasPathSum(root.right, sum-root.val)
        return self.hasPathSum(root.right, sum-root.val) or self.hasPathSum(root.left, sum-root.val)

2：先找到所有路径的各自总和，然后判断给定的总和是否存在

class Solution(object):
    def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root:
            return False
        if sum in self.allPathesSum(root):
            return True
        else:
            return False
    def allPathesSum(self, root):
        if not root.left and not root.right:   #递归结束的标志
            return [root.val]
        elif not root.right:
            tempList = self.allPathesSum(root.left)
        elif not root.left:
            tempList = self.allPathesSum(root.right)
        else:
            left = self.allPathesSum(root.left)
            right = self.allPathesSum(root.right)
            tempList = left + right
        for i in range(len(tempList)):     #累加该结点的值
                tempList[i] += root.val
        return tempList

3：迭代

def hasPathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: bool
        """
        if not root:
            return False
        allPathesSum = [[root,root.val]]
        while allPathesSum:
            curNode = allPathesSum.pop()
            if not curNode[0].left and not curNode[0].right:
                if curNode[1]==sum:
                    return True
            elif not curNode[0].left:
                allPathesSum.insert(0,[curNode[0].right,curNode[0].right.val+curNode[1]])
            elif not curNode[0].right:
                allPathesSum.insert(0,[curNode[0].left,curNode[0].left.val+curNode[1]])
            else:
                allPathesSum.insert(0,[curNode[0].left,curNode[0].left.val+curNode[1]])
                allPathesSum.insert(0,[curNode[0].right,curNode[0].right.val+curNode[1]])
        return False

算法题来自：https://leetcode-cn.com/problems/path-sum/description/