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二叉树的中序遍历

时间:2019-11-24 13:33:01      阅读:73      评论:0      收藏:0      [点我收藏+]

标签:__init__   show   def   交互   https   int   event   exti   node   

题目:给定一个二叉树,返回它的中序 遍历。

来源:https://leetcode-cn.com/problems/binary-tree-inorder-traversal/

法一:网上的代码

思路:利用栈的递归,对每次取的节点进行标记,第一次遍历该节点时,标记为灰色,左子树和右子树标记为白色,注意入栈的顺序是右 中 左,因为最后存储的顺序的左 中 右.

技术图片
# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
class Solution:
    def inorderTraversal(self, root: TreeNode):
        print(root is:)
        print(root)
        WHITE, GRAY = 0, 1
        res = []
        stack = [(WHITE, root)]
        while stack:
            color, node = stack.pop()
            # 如果节点为空则结束本次循环,继续从栈从取元素
            if node is None: continue
            if color == WHITE:
                # 注意这里为了实现中序遍历,入栈的顺序是右 中 左
                stack.append((WHITE, node.right))
                stack.append((GRAY, node))
                stack.append((WHITE, node.left))
            # 如果不为白色,说明已经被遍历过了,则放入res中
            else:
                res.append(node.val)
        return res
# 将list转化为二叉树
# 这里是用队列实现了一个层序遍历,即将list从左往右一层一层依次填充二叉树
def stringToTreeNode(input):
    input = input.strip()
    input = input[1:-1]
    if not input:
        return None
    inputValues = [s.strip() for s in input.split(,)]
    root = TreeNode(int(inputValues[0]))
    nodeQueue = [root]
    front = 0
    index = 1
    while index < len(inputValues):
        node = nodeQueue[front]
        # front指向的是队列的头,即出队列的
        front = front + 1
        item = inputValues[index]
        index = index + 1
        if item != "null":
            leftNumber = int(item)
            node.left = TreeNode(leftNumber)
            nodeQueue.append(node.left)
        if index >= len(inputValues):
            break
        item = inputValues[index]
        index = index + 1
        if item != "null":
            rightNumber = int(item)
            node.right = TreeNode(rightNumber)
            nodeQueue.append(node.right)
    return root
import json
def integerListToString(nums, len_of_list=None):
    if not len_of_list:
        len_of_list = len(nums)
    return json.dumps(nums[:len_of_list])
def main():
    import sys
    import io
    # 实现了交互功能
    def readlines():
        for line in io.TextIOWrapper(sys.stdin.buffer, encoding=utf-8):
            yield line.strip(\n)
    lines = readlines()
    while True:
        try:
            line = next(lines)
            root = stringToTreeNode(line);
            ret = Solution().inorderTraversal(root)
            out = integerListToString(ret);
        except StopIteration:
            break
if __name__ == __main__:
    main()
View Code

 

二叉树的中序遍历

标签:__init__   show   def   交互   https   int   event   exti   node   

原文地址:https://www.cnblogs.com/xxswkl/p/11922021.html

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