二叉树（剑指offer）

目录

1. 前序+ 中序 重建二叉树
2. 二叉树中和为某一值的路径

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1. 前序+ 中序 重建二叉树

前序 ： 根左右
中序： 左根右
从前序知道根节点，在中序中找到那个点（dic)
递归左右

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution(object):
    def buildTree(self, preorder, inorder):
        """
        :type preorder: List[int]
        :type inorder: List[int]
        :rtype: TreeNode
        """
        self.dic = {}
        self.preorder = preorder
        for i, value in enumerate(inorder):
            self.dic[value] = i
        return self.dfs(0, 0, len(inorder) - 1)
        
    def dfs(self, pre_left, in_left, in_right):
        if in_left > in_right:
            return 
        root_value = self.preorder[pre_left]
        root = TreeNode(root_value)
        k = self.dic[root_value]
        left_length = k - in_left
        root.left = self.dfs(pre_left + 1, in_left, k - 1)
        root.right = self.dfs(pre_left + 1 + left_length, k + 1, in_right)
        return root

2. 二叉树中和为某一值的路径

树的DFS

从上往下遍历一遍
当到叶子节点的时候就判断一下是否满足，如果满足就记录下来
判断叶节点： 左右儿子都是空的

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

class Solution(object):
    def findPath(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        self.res = []
        self.path = []
        self.dfs(root, sum)
        return self.res
        
    def dfs(self, root, sum):
        if not root:
            return 
        self.path.append(root.val)
        sum -= root.val
        if not root.left and not root.right and not sum:
            self.res.append(list(self.path))
        self.dfs(root.left, sum)
        self.dfs(root.right, sum)
        self.path.pop()