二叉树的遍历

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)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def postorderTraversal(self, root: TreeNode) -> List[int]:
        if root == None:
            return []
        pre = None
        res = []
        stack = [root]
        while stack:
            temp = stack[-1]  # 这里不能直接用pop,因为只有读取该节点的时候才要已经读取了的节点pop出栈
            if (temp.left == None and temp.right == None) or (pre and (pre == temp.left or pre == temp.right)):
            # 用pre来进行标志,由于是后序遍历,
            # 如果当前节点是叶节点(左右子节点为空),读取值,
            # 如果不是叶节点但是pre是他的子节点,
            # 就说明这个节点也是当前应该读取的点
                res.append(temp.val)
                pre = temp
                stack.pop()
            else:  # 因为栈先进后出,因此子节点压入的顺序是先右后左。
                if temp.right:
                    stack.append(temp.right)
                if temp.left:
                    stack.append(temp.left)
        return res

4.按层遍历:从上到下、从左到右按层遍历

# 先进先出选用队列结构
import queue
def layerTraverse(head):
    if not head:
        return None
    que = queue.Queue()      # 创建先进先出队列
    que.put(head)
    while not que.empty():
        head = que.get()    # 弹出第一个元素并打印
        print(head.val)
        if head.left:       # 若该节点存在左子节点,则加入队列(先push左节点)
            que.put(head.left)
        if head.right:      # 若该节点存在右子节点,则加入队列(再push右节点)
            que.put(head.right)

 

参考博客:https://www.cnblogs.com/icekx/p/9127569.html

 

posted @ 2020-04-02 20:38  USTC丶ZCC  阅读(188)  评论(0编辑  收藏  举报