二叉树先序、中序、后序遍历 递归与非递归 Python实现
1.先序遍历:根节点->左子树->右子树
1 # 先序打印二叉树(递归) 2 def preOrderTraverse(node): 3 if node is None: 4 return None 5 print(node.val) 6 preOrderTraverse(node.left) 7 preOrderTraverse(node.right)
1 # 先序打印二叉树(非递归) 2 def preOrderTravese(node): 3 stack = [node] 4 while len(stack) > 0: 5 print(node.val) 6 if node.right is not None: 7 stack.append(node.right) 8 if node.left is not None: 9 stack.append(node.left) 10 node = stack.pop()
2.中序遍历:左子树->根节点->右子树
1 # 中序打印二叉树(递归) 2 def inOrderTraverse(node): 3 if node is None: 4 return None 5 inOrderTraverse(node.left) 6 print(node.val) 7 inOrderTraverse(node.right)
1 # 中序打印二叉树(非递归) 2 def inOrderTraverse(node): 3 stack = [] 4 pos = node 5 while pos is not None or len(stack) > 0: 6 if pos is not None: 7 stack.append(pos) 8 pos = pos.left 9 else: 10 pos = stack.pop() 11 print(pos.val) 12 pos = pos.right
3.后序遍历:左子树->右子树->根节点
1 # 后序打印二叉树(递归) 2 def postOrderTraverse(node): 3 if node is None: 4 return None 5 postOrderTraverse(node.left) 6 postOrderTraverse(node.right) 7 print(node.val)
1 # 后序打印二叉树(非递归) 2 # 使用两个栈结构 3 # 第一个栈进栈顺序:左节点->右节点->跟节点 4 # 第一个栈弹出顺序: 跟节点->右节点->左节点(先序遍历栈弹出顺序:跟->左->右) 5 # 第二个栈存储为第一个栈的每个弹出依次进栈 6 # 最后第二个栈依次出栈 7 def postOrderTraverse(node): 8 stack = [node] 9 stack2 = [] 10 while len(stack) > 0: 11 node = stack.pop() 12 stack2.append(node) 13 if node.left is not None: 14 stack.append(node.left) 15 if node.right is not None: 16 stack.append(node.right) 17 while len(stack2) > 0: 18 print(stack2.pop().val)
4.按层遍历:从上到下、从左到右按层遍历
1 # 先进先出选用队列结构 2 import queue 3 def layerTraverse(head): 4 if not head: 5 return None 6 que = queue.Queue() # 创建先进先出队列 7 que.put(head) 8 while not que.empty(): 9 head = que.get() # 弹出第一个元素并打印 10 print(head.val) 11 if head.left: # 若该节点存在左子节点,则加入队列(先push左节点) 12 que.put(head.left) 13 if head.right: # 若该节点存在右子节点,则加入队列(再push右节点) 14 que.put(head.right)
5.二叉树节点个数
1 # 求二叉树节点个数 2 def treeNodenums(node): 3 if node is None: 4 return 0 5 nums = treeNodenums(node.left) 6 nums += treeNodenums(node.right) 7 return nums + 1
6.二叉树的最大深度
1 # 二叉树的最大深度 2 def bTreeDepth(node): 3 if node is None: 4 return 0 5 ldepth = bTreeDepth(node.left) 6 rdepth = bTreeDepth(node.right) 7 return (max(ldepth, rdepth) + 1)
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