二叉树（1）-----遍历

时间/空间复杂度分析

时间复杂度均为O(n)；空间复杂度平均情况下为O(logn)，最坏情况下为O(n).

Morris遍历：时间为O（n），空间为O（1）

一、前序遍历：

递归方式：

def preorder(tree):
    if tree:
        print(tree.val)
        preorder(tree.getLeftChild())
        preorder(tree.getRightChild())

 

非递归方式：时间复杂度O（n)，空间复杂度O（n）

 

def preOrder(head):
    if not head:
        return None
    res ,stack = [] , [head]
    while stack:
        cur = stack.pop()
        res.append(cur.val)
        if cur.right:
            stack.append(cur.right)
        if cur.left:
            stack.append(cur.left)
    return res

二、中序遍历：

递归方式：

def inorder(tree):    
    if tree:
        inorder(tree.left)
        print(tree.val)
        inorder(tree.right)

非递归方式：

 

def InOrder(head):
    if not head:
        return None
    res ,stack = [] , []
    cur = head
    while stack or cur:
        if cur:
            stack.append(cur)
            cur = cur.left
        else:
            cur = stack.pop()
            res.append(cur.val)
            cur =cur.right
    return res

 

 三、后序遍历：

递归

 

def postorder(tree):
    if tree:
        postorder(tree.left)
        postorder(tree.right))
        print(tree.val)

 

非递归：

def PosOrder(head):
    if not head:
        return []
    s1 , s2 = [head] , []
    cur = head
    while s1:
        cur = s1.pop()
        s2.append(cur.val)
        if cur.left:
            s1.append(cur.left)
        if cur.right:
            s1.append(cur.right)
    return s2[::-1] 

 

 

class Node:
    def __init__(self,value):
        self.val = value
        self.left = None
        self.right = None
def posOrder(head):
    if not head:
        return []
    res = []
    cur = [head]
    while cur:
        node = cur[-1]
        if node.left and node.left != head and node.right!= head:
            cur.append(node.left)
        elif node.right and node.right != head:
            cur.append(node.right)
        else:
            res.append(cur.pop().val)
            head = node
    return res

if __name__ == '__main__':
    tree = Node(5)
    tree.left = Node(4)
    tree.right = Node(3)
    tree.left.left = Node(2)
    tree.left.right = Node(1)
    posOrder(tree)

 

四、层次遍历：

两个栈：

 

一个栈：

class Node: def __init__(self,value): self.val = value self.left = None self.right = None def posOrder(root): if not root: return [] last = root nlast = None stack = [root] res = [[]] while stack: head = stack.pop(0) res[-1].append(head.val) if head.left: stack.append(head.left) nlast = head.left if head.right: stack.append(head.right) nlast = head.right if head == last and stack: last = nlast res.append([]) return res if __name__ == '__main__': tree = Node(5) tree.left = Node(4) tree.right = Node(3) tree.left.left = Node(2) tree.left.right = Node(1) posOrder(tree)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0个栈：（DFS）求每层的平均值

#####求每层的平均值   
 def averageOfLevels(self, root):
        """
        :type root: TreeNode
        :rtype: List[float]
        """
        def computeSum(root,height,sumlist,countlist):
            if not root:
                return 0
            if height>=len(countlist):
                countlist.append(0)
                sumlist.append(0)
            sumlist[height]+=root.val
            countlist[height]+=1.0
            computeSum(root.left,height+1,sumlist,countlist)
            computeSum(root.right,height+1,sumlist,countlist)
        sumlist=[]
        countlist=[]
        computeSum(root,0,sumlist,countlist)
        return [i/j if j!=0.0 else 0.0 for i,j in zip(sumlist,countlist)]