105从前序与中序遍历序列构造二叉树

题目: 根据一棵树的前序遍历与中序遍历构造二叉树。

来源: https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/

总结: 法一的解法只适用于前序中序,后序中序.不适用于前序后序,法二解法对这三种都适用,缺点是内存消耗大.

法一: 官方代码

思路: 利用分治和递归,对前序遍历元素进行遍历,再对中序遍历进行递归回溯,关键是要观察出两个遍历的结构特点,以及递归的写法.

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
from typing import List
class Solution:
    def buildTree(self, preorder, inorder):
        # global pre_idx
        def helper(in_left=0, in_right=len(inorder)):
            nonlocal pre_idx
            # global pre_idx
            # 如果左
            if in_left == in_right:
                return None
            # 先找出前序遍历索引对应的数
            root_val = preorder[pre_idx]
            # 再将其建一个节点
            root = TreeNode(root_val)
            # 求出中序遍历中该元素对应的索引
            index = idx_map[root_val]
            # 索引加一
            pre_idx += 1
            # 寻找节点的左节点,(in_left, index)是左子树所有的节点值
            root.left = helper(in_left, index)
            # 寻找节点的右节点
            root.right = helper(index + 1, in_right)
            return root
        # 对中序遍历建立一个字典,key是inorder中的值,value是值对应的索引
        idx_map = {val: idx for idx, val in enumerate(inorder)}
        pre_idx = 0
        return helper()
if __name__ == '__main__':
    def list_to_binarytree(a):
        if a[0] is None:
            return []
        stack = []
        root = TreeNode(a.pop(0))
        stack.append(root)
        while a:
            # 一层一层的遍历
            stack_next = stack
            stack = []
            # a有可能提前变为[],这时直接结束内循环
            while stack_next and a:
                # 注意这里的b不能是空对象的地址,否则会出错
                b = stack_next.pop(0)
                # 为了保证结构的统一特别引入了555,stack中555表示这里是一个空节点,
                # stack中只有两种值一种是555,另一种是节点的地址
                # 上一层的空节点对应下一层a中的两个None,依次弹出
                if b == 555:
                    a.pop(0)
                    a.pop(0)
                    continue
                # 如果某个节点为None,则将其地址标记为555
                if a[0] == None:
                    a.pop(0)
                    stack.append(555)
                else:
                    b.left = TreeNode(a.pop(0))
                    stack.append(b.left)
                if a[0] == None:
                    a.pop(0)
                    stack.append(555)
                else:
                    b.right = TreeNode(a.pop(0))
                    stack.append(b.right)
    duixiang = Solution()
    a = [1,
         2, 3,
         4, 5, 6, 7,
         8, 9, 10, None, None, None, 11, 12,
         None, None, None, None,
         None, None, None, None,
         None, None, None, None, 13, 14]
    preorder = [1, 2, 4, 8, 9, 5, 10, 3, 6, 7, 11, 13, 14, 12]
    inorder = [8, 4, 9, 2, 10, 5, 1, 6, 3, 13, 11, 14, 7, 12]
    root = list_to_binarytree(a)
    a = duixiang.buildTree(preorder, inorder)

法二: 模仿889的写法,每次取前序遍历的第一个元素,缺点是内存消耗大

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
# 执行用时 :196 ms, 在所有 python3 提交中击败了70.59% 的用户
# 内存消耗 :87.3 MB, 在所有 python3 提交中击败了72.23%的用户
# 模仿889由前序和后序构造二叉树的写法
class Solution:
    def buildTree(self, preorder, inorder):
        def recursion(preorder=preorder, inorder= inorder):
            if len(inorder) == 0: return None
            root = TreeNode(preorder[0])
            if len(preorder) == 1 : return root
            L = inorder.index(root.val)
            root.left = recursion(preorder[1:L+1], inorder[:L])
            root.right = recursion(preorder[L+1:], inorder[L+1:])
            return root
        return  recursion()
if __name__ == '__main__':
    def list_to_binarytree(a):
        if a[0] is None:
            return []
        stack = []
        root = TreeNode(a.pop(0))
        stack.append(root)
        while a:
            # 一层一层的遍历
            stack_next = stack
            stack = []
            # a有可能提前变为[],这时直接结束内循环
            while stack_next and a:
                # 注意这里的b不能是空对象的地址,否则会出错
                b = stack_next.pop(0)
                # 为了保证结构的统一特别引入了555,stack中555表示这里是一个空节点,
                # stack中只有两种值一种是555,另一种是节点的地址
                # 上一层的空节点对应下一层a中的两个None,依次弹出
                if b == 555:
                    a.pop(0)
                    a.pop(0)
                    continue
                # 如果某个节点为None,则将其地址标记为555
                if a[0] == None:
                    a.pop(0)
                    stack.append(555)
                else:
                    b.left = TreeNode(a.pop(0))
                    stack.append(b.left)
                if a[0] == None:
                    a.pop(0)
                    stack.append(555)
                else:
                    b.right = TreeNode(a.pop(0))
                    stack.append(b.right)
    duixiang = Solution()
    a = [1,
         2, 3,
         4, 5, 6, 7,
         8, 9, 10, None, None, None, 11, 12,
         None, None, None, None,
         None, None, None, None,
         None, None, None, None, 13, 14]
    preorder = [1, 2, 4, 8, 9, 5, 10, 3, 6, 7, 11, 13, 14, 12]
    inorder = [8, 4, 9, 2, 10, 5, 1, 6, 3, 13, 11, 14, 7, 12]
    root = list_to_binarytree(a)
    a = duixiang.buildTree(preorder, inorder)
    print(a)