Leetcode刷题笔记-Tree

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105. Construct Binary Tree from Preorder and Inorder Traversal

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

# 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
        """
        if inorder:
            ind = inorder.index(preorder.pop(0))
            root = TreeNode(inorder[ind])
            root.left = self.buildTree(preorder, inorder[0: ind])
            root.right = self.buildTree(preorder, inorder[ind+1:])
            return root

106. Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

class Solution(object):
    def buildTree(self, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if not postorder or not inorder:
            return
        
        root = TreeNode(postorder.pop())
        inorderIndex = inorder.index(root.val)
        
        root.right = self.buildTree(inorder[inorderIndex+1:], postorder)
        root.left = self.buildTree(inorder[:inorderIndex], postorder)
        
        return root

230. Kth Smallest Element in a BST

 递归访问左节点->自己->右节点

class Solution(object):
    def kthSmallest(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: int
        """
        self.count = self.result = 0
        self.helper(root, k)
        return self.result
    
    def helper(self, root, k):
        if not root:
            return  
        
        self.helper(root.left, k)
        self.count += 1
        
        if self.count == k:
            self.result = root.val
            return
        
        self.helper(root.right, k)

94. Binary Tree Inorder Traversal

inorder preorder的可以整理一下

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        
        return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right)

236. Lowest Common Ancestor of a Binary Tree

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        stack = [root]
        parent = {root: None}

        while p not in parent or q not in parent:
            node = stack.pop()
            if node.left:
                parent[node.left] = node
                stack.append(node.left)
            if node.right:
                parent[node.right] = node
                stack.append(node.right)
            
        ancestors = set()
        while p:
            ancestors.add(p)
            p = parent[p]
        
        while q not in ancestors:
            q = parent[q]
            
        return q