python遍歷二叉樹
阿新 • • 發佈:2018-10-23
elf class [] res 兩個棧 樹節點 roo pri pytho
定義二叉樹:
class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None
構建二叉樹:
# 返回構造的TreeNode根節點 def reConstructBinaryTree(self, pre, tin): if not pre or not tin: return None root = TreeNode(pre[0])#根節點 # 判斷輸入的兩個序列是不是匹配if set(pre) != set(tin): return None i = tin.index(root.val) # i == 3 root.left = self.reConstructBinaryTree(pre[1:i+1],tin[:i]) # 列表:左閉右開 root.right = self.reConstructBinaryTree(pre[i+1:],tin[i+1:]) return root
BFS:
def BFS(self, root): #寬度優先遍歷BFS array = [] result = [] if root == None: return result array.append(root) while array: newNode = array.pop(0) # 根結點 result.append(newNode.val) if newNode.left != None: array.append(newNode.left) if newNode.right != None: array.append(newNode.right)return result
先序遍歷:
1.遞歸版本:
def pre_traversal(self): ret = [] def traversal(head): if not head: return ret.append(head.val) traversal(head.left) traversal(head.right) traversal(self.root) return ret
2.非遞歸版本
# 先序打印二叉樹(非遞歸) def preOrderTravese(node): stack = [node] while len(stack) > 0: print(node.val) if node.right is not None: stack.append(node.right) if node.left is not None: stack.append(node.left) node = stack.pop()
中序遍歷:
1.遞歸版本
def in_traversal(self): ret = [] def traversal(head): if not head: return traversal(head.left) ret.append(head.val) traversal(head.right) traversal(self.root) return ret
2.非遞歸版本
# 中序打印二叉樹(非遞歸) def inOrderTraverse(node): stack = [] pos = node while pos is not None or len(stack) > 0: if pos is not None: stack.append(pos) pos = pos.left else: pos = stack.pop() print(pos.val) pos = pos.right
後序遍歷:
1.遞歸版本
def post_traversal(self): ret = [] def traversal(head): if not head: return traversal(head.left) traversal(head.right) ret.append(head.val) traversal(self.root) return ret
2.非遞歸版本
# 後序打印二叉樹(非遞歸) # 使用兩個棧結構 # 第一個棧進棧順序:左節點->右節點->跟節點 # 第一個棧彈出順序: 跟節點->右節點->左節點(先序遍歷棧彈出順序:跟->左->右) # 第二個棧存儲為第一個棧的每個彈出依次進棧 # 最後第二個棧依次出棧 def postOrderTraverse(node): stack = [node] stack2 = [] while len(stack) > 0: node = stack.pop() stack2.append(node) if node.left is not None: stack.append(node.left) if node.right is not None: stack.append(node.right) while len(stack2) > 0: print(stack2.pop().val)
求二叉樹最大深度:
# 二叉樹的最大深度 def bTreeDepth(node): if node is None: return 0 ldepth = bTreeDepth(node.left) rdepth = bTreeDepth(node.right) return (max(ldepth, rdepth) + 1)
求二叉樹節點個數:
# 求二叉樹節點個數 def treeNodenums(node): if node is None: return 0 nums = treeNodenums(node.left) nums += treeNodenums(node.right) return nums + 1
python遍歷二叉樹