513.找樹左下角的值
迭代法比較簡單，層序遍歷，找到最下面一層的第一個節點。題目已經說明節點數>=1了
class Solution:def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:queue = collections.deque()queue.append(root)result = root.valwhile queue:size = len(queue)for i in range(size):node = queue.popleft()if i == 0:result = node.valif node.left:queue.append(node.left)if node.right:queue.append(node.right)return result遞歸法
題解中遇到葉子節點并且當前深度比最大深度更大時更換結果值，但是最深的節點必定是葉子節點，所以不必判斷是葉子節點
class Solution:def __init__(self):self.result = 0self.max_depth = 0def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:self.getValue(root, 1)return self.resultdef getValue(self, node, depth):if not node:returnif depth > self.max_depth:self.max_depth = depthself.result = node.valself.getValue(node.left, depth+1)self.getValue(node.right, depth+1)112. 路徑總和
class Solution:def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:if not root:return Falsereturn self.traversal(root, 0, targetSum)def traversal(self, node, cur_sum, target_sum):if not node:return Falsecur_sum += node.valif not node.left and not node.right:if cur_sum == target_sum:return Truereturn self.traversal(node.left, cur_sum, target_sum) or self.traversal(node.right, cur_sum, target_sum)下面的代碼思路更清晰
不要去累加然后判斷是否等于目標和，那么代碼比較麻煩，可以用遞減，讓計數器count初始為目標和，然后每次減去遍歷路徑節點上的數值。
class Solution:def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:if not root:return Falsereturn self.traversal(root, targetSum-root.val)def traversal(self, node, count):if not node.left and not node.right:return count == 0if node.left:if self.traversal(node.left, count-node.left.val):return Trueif node.right:if self.traversal(node.right, count-node.right.val):return Truereturn False路徑總和：返回是否存在路徑
路徑總和II：返回滿足條件的所有路徑
下面為路徑總和II的代碼
class Solution:def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:result = []if not root:return resultself.getPath(root, [root.val], result, targetSum-root.val)return resultdef getPath(self, node, path, result, count):if not node.left and not node.right:if count == 0:result.append(path[:])if node.left:path.append(node.left.val)self.getPath(node.left, path, result, count-node.left.val)path.pop()if node.right:path.append(node.right.val)self.getPath(node.right, path, result, count-node.right.val)path.pop()106.從中序與后序遍歷序列構造二叉樹
下面為每層遞歸定義了新的vector（就是數組），既耗時又耗空間。可以使用索引的方式，每次確定區間的左右索引
class Solution:def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:if len(inorder) <= 0:returnvalue = postorder[-1]index = inorder.index(value)left = self.buildTree(inorder[0:index], postorder[0:index])right = self.buildTree(inorder[index+1:], postorder[index:-1])return TreeNode(value, left, right)105. 從前序與中序遍歷序列構造二叉樹
class Solution:def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:if not preorder:returnvalue = preorder[0]index = inorder.index(value)left = self.buildTree(preorder[1:index+1], inorder[0:index])right = self.buildTree(preorder[index+1:], inorder[index+1:])return TreeNode(value, left, right)
