【LeetCode with Python】 Path Sum

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原题页面：https://oj.leetcode.com/problems/path-sum/

题目类型：递归迭代，搜索剪枝

难度评价：★

本文地址：/article/1377522.html

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.

For 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.

简单的递归算法。但是注意一点，当树为空树时，无论sum是多少都一定要返回False，即使此时sum为0。另外因为只要确定有解即可，因此如果某次迭代发现left已经找到解，那么就不需要再去右子树中寻找解，即及时剪枝。

但是，因为可能有负数出现，所以不能因为path sum已经大于指定sum值就认为该分支可以剪出。

class Solution:

    def doHasPathSum(self, root, total, sum):
        if None == root.left and None == root.right:
            return sum == (total + root.val)

        left= False
        right = False
        if None != root.left:
            left = self.doHasPathSum(root.left, total + root.val, sum)
        if False != left and None != root.right:     # pruning
            right = self.doHasPathSum(root.right, total + root.val, sum)
        return left or right

    # @param root, a tree node
    # @param sum, an integer
    # @return a boolean
    def hasPathSum(self, root, sum):
        if None == root:
            return False ###
        return self.doHasPathSum(root, 0, sum)