二叉树的中序遍历

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.

# 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

来源：https://www.cnblogs.com/xxswkl/p/11922021.html

标签

中序遍历