算法练习Day16 （Leetcode/Python-二叉树）
递归何时需要返回值
1）搜索一整棵树且不需要处理递归返回值，就不需要返回值
2）需要搜索一整棵树且需要处理递归返回值，则需要返回
3）搜索其中一条符合条件的路径，就需要返回值，以便在遇到合适的路径时返回。
112. Path Sum
Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.
A leaf is a node with no children.
思路一：
使用递归法。使用DFS深度优先，尝试探索一条路径直达一个叶子节点，如果该路径不满足要求，则回溯以探索其他路径。
注意：这里用递减的方式记录该路径的节点值的sum直到减为0使得代码更加简洁。注意递归代码里是在input时就算好了当前路径（包含当前节点的子节点）的sum值，进入递归代码后先判断sum是否符合要求。如果不符合，跳出递归后把当前节点的子节点减去，然后再探索下一个子节点。
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution(object):
def traversal(self, cur, count):
# reach the leave node and achieve the targetSum, return True
if not cur.left and not cur.right and count == 0:
return True
if not cur.left and not cur.right:
return False
if cur.left:
count -= cur.left.val
if self.traversal(cur.left, count):
return True
count += cur.left.val # backtrack
# 如果当前的子节点加上后未出现满足要求的路径，就减去当前子节点的值，回归原状以探索下一个路径。
if cur.right:
count -= cur.right.val
if self.traversal(cur.right, count):
return True
count += cur.right.val # backtrack
return False
def hasPathSum(self, root, targetSum):
"""
:type root: TreeNode
:type targetSum: int
:rtype: bool
"""
if not root:
return False
return self.traversal(root, targetSum-root.val) # 记得减去root的值啊！！！这种递归写法把当前路径的sum作为递归函数的input，隐式地实现了回溯。写法更加简洁。
class Solution(object):
def hasPathSum(self, root, targetSum):
"""
:type root: TreeNode
:type targetSum: int
:rtype: bool
"""
if not root:
return False
if not root.left and not root.right and targetSum == root.val:
return True
return self.hasPathSum(root.left, targetSum-root.val) or self.hasPathSum(root.right, targetSum-root.val)思路二：迭代法。 这也是DFS。
相较于递归法，迭代法通常在内存上更有效率，因为递归会在调用栈上增加额外的内存负担。有时语言或环境优化不够的情况下，迭代会更有效率。迭代法代码可能更容易理解。
递归法则是代码更为简洁。有时也更利于维护。
class Solution(object):
# Iteration
def hasPathSum(self, root, targetSum):
"""
:type root: TreeNode
:type targetSum: int
:rtype: bool
"""
if not root:
return False
st = [(root, root.val)]
while st:
node, path_sum = st.pop()
if not node.left and not node.right and path_sum == targetSum:
return True
if node.right:
st.append((node.right, path_sum+node.right.val))
if node.left:
st.append((node.left, path_sum+node.left.val))
return False相比上一道题，这里需要把所有满足要求的path记录并输出。所以就不用在traversal中提前return，而是再出现满足要求的path后，加入result。并且由于要记录所有的path，要记得回溯的时候pop出刚加进来的叶节点。
class Solution(object):
def __init__(self):
self.result = []
self.path = []
def traversal(self, cur, count):
if not cur.left and not cur.right and count == 0:
self.result.append(self.path[:])
return
if not cur.left and not cur.right:
return False
if cur.left: # left
count -= cur.left.val
self.path.append(cur.left.val)
self.traversal(cur.left, count)
count += cur.left.val
self.path.pop()
if cur.right: # right
count -= cur.right.val
self.path.append(cur.right.val)
self.traversal(cur.right, count)
count += cur.right.val
self.path.pop()
return
def pathSum(self, root, targetSum):
"""
:type root: TreeNode
:type targetSum: int
:rtype: List[List[int]]
"""
self.result = []
self.path = []
if not root:
return self.result
self.path.append(root.val)
self.traversal(root, targetSum-root.val)
return self.resultclass Solution:
def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
if not root:
return []
stack = [(root, [root.val])]
res = []
while stack:
node, path = stack.pop()
if not node.left and not node.right and sum(path) == targetSum:
res.append(path)
if node.right:
stack.append((node.right, path + [node.right.val]))
if node.left:
stack.append((node.left, path + [node.left.val]))
return res
404. Sum of Left Leaves
Given the root of a binary tree, return the sum of all left leaves.
A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.
思路：首先明确什么是左叶子节点。即该节点无左右子节点，且是父节点的左节点。右子节点可能有它的左叶子节点，左子节点也可能有它的左叶子节点。注意，要通过父节点来判断是否是左叶子节点。
递归三要素：1） 递归函数参数和返回值：root和数值之和
2）终止条件：
if root is None: return 0;
if root.left is None && root.right is None: return 0;3) 单层递归的逻辑：
遇到左叶子节点的时候，记下数值。通过递归法求左子树左叶子之和和右子树左叶子之和，相加起来就是整个树的左叶子之和。
递归法：
class Solution(object):
def sumOfLeftLeaves(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
if not root.left and not root.right:
return 0
leftValue = self.sumOfLeftLeaves(root.left) # left
if root.left and not root.left.left and not root.left.right: #The left node is a left leave.
leftValue = root.left.val
rightValue = self.sumOfLeftLeaves(root.right) # right
sum_val = leftValue + rightValue # middle
return sum_val 递归法简化版：
class Solution(object):
def sumOfLeftLeaves(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root is None:
return 0
leftValue = 0
if root.left is not None and root.left.left is None and root.left.right is None:
leftValue = root.left.val
return leftValue + self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right)
迭代法：
class Solution(object):
def sumOfLeftLeaves(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root is None:
return 0
st = [root]
result = 0
while st:
node = st.pop()
if node.left and node.left.left is None and node.left.right is None:
result += node.left.val
if node.right:
st.append(node.right)
if node.left:
st.append(node.left)
return result Reference：
代码随想录