Leetcode 332. Reconstruct Itinerary

Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], reconstruct the itinerary in order. All of the tickets belong to a man who departs from JFK. Thus, the itinerary must begin with JFK.

Note:

1. If there are multiple valid itineraries, you should return the itinerary that has the smallest lexical order when read as a single string. For example, the itinerary ["JFK", "LGA"] has a smaller lexical order than ["JFK", "LGB"].
2. All airports are represented by three capital letters (IATA code).
3. You may assume all tickets form at least one valid itinerary.

Example 1:

Input: [["MUC", "LHR"], ["JFK", "MUC"], ["SFO", "SJC"], ["LHR", "SFO"]]
Output: ["JFK", "MUC", "LHR", "SFO", "SJC"]

Example 2:

Input: [["JFK","SFO"],["JFK","ATL"],["SFO","ATL"],["ATL","JFK"],["ATL","SFO"]]
Output: ["JFK","ATL","JFK","SFO","ATL","SFO"]
Explanation: Another possible reconstruction is ["JFK","SFO","ATL","JFK","ATL","SFO"].
             But it is larger in lexical order.

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这题的特殊之处在于，深度优先遍历，参考下图，优先深下去，然后字典序优先的靠前。可以从先序的角度去思考，先序到KUL卡住了，看看另外一棵子树是不是能遍历完，这就是后序的思路了。

从discussion学的别人代码：

class Solution:
    def findItineraryNoneRecursive(self, tickets):
        d = {}
        tickets.sort(key=lambda x: x[1], reverse=True)

        for ticket in tickets:
            fro, to = ticket[0], ticket[1]
            if (fro not in d):
                d[fro] = [to]
            elif (fro in d):
                d[fro].append(to)

        revert,sta = [],["JFK"]
        while (sta):
            while(sta[-1] in d and d[sta[-1]]):
                top = d[sta[-1]].pop()
                sta.append(top)
            revert.append(sta.pop())
        return revert[::-1]

    def findItinerary(self, tickets):
        d = {}
        tickets.sort(key=lambda x: x[1], reverse=True)

        for ticket in tickets:
            fro, to = ticket[0], ticket[1]
            if (fro not in d):
                d[fro] = [to]
            elif (fro in d):
                d[fro].append(to)

        revert,sta = [],["JFK"]
        self.dfs("JFK", d, revert)
        return revert[::-1]
    def dfs(self, cur, d, revert):
        while (cur in d and d[cur]):
            self.dfs(d[cur].pop(), d, revert)
        revert.append(cur)

s = Solution()
print(s.findItinerary([["JFK","KUL"],["JFK","NRT"],["NRT","JFK"]])) #['JFK', 'NRT', 'JFK', 'KUL']
#print(s.findItinerary([["JFK","KUL"],["JFK","NRT"],["KUL","JFK"]])) #['JFK', 'KUL', 'JFK', 'NRT']
#print(s.findItinerary([["JFK","KUL"],["JFK","NRT"],["KUL","JFK"],["NRT","JFK"]])) #['JFK', 'KUL', 'JFK', 'NRT', 'JFK']

后序遍历的思路，这道题学到了一种新玩法post3：

from itertools import permutations


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


class Solution:
    def p1(self, root):
        if (root == None):
            return
        print(root.val)
        self.p1(root.left)
        self.p1(root.right)

    def p2(self, root):
        stack = []
        while (root != None or stack):
            while (root != None):
                print(root.val)
                stack.append(root)
                root = root.left
            if (stack):
                root = stack.pop()
                root = None if root == None else root.right

    def i1(self, root):
        if (root == None):
            return
        self.i1(root.left)
        print(root.val)
        self.i1(root.right)

    def i2(self, root):
        stack = []
        while (root != None or stack):
            while (root != None):
                stack.append({'node': root, 'tag': 'l'})
                root = root.left
            if (stack):
                cur = stack.pop()
                if (cur["tag"] == 'l'):
                    print(cur["node"].val)
                    if (cur["node"].right != None):
                        stack.append({'node': cur["node"].right, 'tag': 'r'})
                    root = None
                elif (cur["tag"] == 'r'):
                    root = cur["node"]

    # i3 is better, inorder doesn't need extra flg in fact
    def i3(self, root):
        stack = []
        while (root != None or stack):
            while (root != None):
                stack.append(root)
                root = root.left
            if (stack):
                root = stack.pop()
                print(root.val)
                root = root.right if (root.right != None) else None

    def post1(self, root):
        if (root == None):
            return
        self.post1(root.left)
        self.post1(root.right)
        print(root.val)

    def post2(self, root):
        stack = []
        while (root != None or stack):
            while (root != None):
                stack.append({'node': root, 'tag': 'l'})
                root = root.left
            if (stack):
                cur = stack.pop()
                if (cur["tag"] == 'l'):
                    stack.append({'node': cur["node"], 'tag': 'r'})
                    root = None if cur["node"].right == None else cur["node"].right
                elif (cur["tag"] == 'r'):
                    print(cur["node"].val)
                    root = None

    def get_not_visited_child(self, parent, visited):
        if (parent.left and parent.left not in visited):
            return parent.left
        if (parent.right and parent.right not in visited):
            return parent.right
        return None

    #新玩法
    def post3(self,root):
        sta = [root]
        visited = {root}
        revert = []
        while (sta):
            next = self.get_not_visited_child(sta[-1], visited)
            while (next):
                sta.append(next)
                visited.add(next)
                next = self.get_not_visited_child(sta[-1], visited)
            if (sta):
                revert.append(sta.pop().val)
        return revert

n1 = TreeNode(1)
n2 = TreeNode(2)
n3 = TreeNode(3)
n4 = TreeNode(4)
n5 = TreeNode(5)
n6 = TreeNode(6)
n7 = TreeNode(7)
n1.left = n2
n1.right = n3
n2.left = n4
n2.right = n5
n3.left = n6
n3.right = n7
s = Solution()
s.post2(n1)
print(s.post3(n1))

更多可以参考https://blog.youkuaiyun.com/taoqick/article/details/82495508