112. Path Sum [easy] (Python)

题目链接

https://leetcode.com/problems/path-sum/

题目原文

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.

题目翻译

给定一个二叉树及一个和数sum，判断这个树是否有一条从根到叶的路径，这条路径上所有数之和为sum。比如，给定sum=22，二叉树为：

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

那么应该返回true，因为存在路径

5->4->11->2

，其和为22。

思路方法

思路一

用深度优先搜索（DFS）遍历所有可能的从根到叶的路径，要注意每深一层要从和中减去相应节点的数值。下面是递归实现的代码。

代码

# 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 not root:
return False
if not root.left and not root.right:
return True if sum == root.val else False
else:
return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)

思路二

DFS的非递归实现，用栈实现。

代码

# 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
"""
stack = [(root, sum)]
while len(stack) > 0:
node, tmp_sum = stack.pop()
if node:
if not node.left and not node.right and node.val == tmp_sum:
return True
stack.append((node.right, tmp_sum-node.val))
stack.append((node.left, tmp_sum-node.val))
return False

思路三

BFS方法，用队列实现。

代码

# 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
"""
queue = [(root, sum)]
while len(queue) > 0:
node, tmp_sum = queue.pop()
if node:
if not node.left and not node.right and node.val == tmp_sum:
return True
queue.insert(0, (node.right, tmp_sum-node.val))
queue.insert(0, (node.left, tmp_sum-node.val))
return False

思路四

如果说上面都是比较常规的方法，那么后序遍历算是比较新奇的解法了。虽然也用的栈，但后序遍历的一大好处是它直接将路径保存在栈中，每次进入不同的层不需要记录当前的和。算是与DFS各有所长吧。

代码

# 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
"""
pre, cur = None, root
tmp_sum = 0
stack = []
while cur or len(stack) > 0:
while cur:
stack.append(cur)
tmp_sum += cur.val
cur = cur.left
cur = stack[-1]
if not cur.left and not cur.right and tmp_sum == sum:
return True
if cur.right and pre != cur.right:
cur = cur.right
else:
pre = cur
stack.pop()
tmp_sum -= cur.val
cur = None
return False

PS: 新手刷LeetCode，新手写博客，写错了或者写的不清楚还请帮忙指出，谢谢！

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