重建二叉树

已知先序遍历和中序遍历及已知中序遍历和后序遍历重建二叉树的Python递归和迭代解法

105. Construct Binary Tree from Preorder and Inorder Traversal

https://leetcode.com/problems/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.

Solution

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

Iteration version

https://discuss.leetcode.com/topic/35826/python-recursion-version-and-iteration-version-easy-to-understand

class Solution(object):
    def buildTree(self, preorder, inorder):
        if len(preorder) == 0:
            return None
            
        head = TreeNode(preorder[0])
        stack = [head]
        i = 1
        j = 0
        
        while i < len(preorder):
            temp = None
            t = TreeNode(preorder[i])
            while stack and stack[-1].val == inorder[j]:
                temp = stack.pop()
                j += 1
            if temp:
                temp.right = t
            else:
                stack[-1].left = t
            stack.append(t)
            i += 1
        
        return head

106. Construct Binary Tree from Inorder and Postorder Traversal

https://leetcode.com/problems/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.

Solution

# 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, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        if inorder:
            temp = postorder[-1]
            postorder.remove(temp)  # duplicates do not exist in the tree
            index = inorder.index(temp)
            root = TreeNode(temp)
            root.right = self.buildTree(inorder[index+1:],postorder)  # 先重建右子树
            root.left = self.buildTree(inorder[:index],postorder)
            
            return root

Iteration version

# 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, inorder, postorder):
        """
        :type inorder: List[int]
        :type postorder: List[int]
        :rtype: TreeNode
        """
        oi, pi = 0,0
        stack = []
        cur = None
        while pi < len(postorder):
            if len(stack) and stack[-1].val == postorder[pi]:
                stack[-1].right = cur
                cur = stack.pop()
                pi += 1
            else:
                stack.append(TreeNode(inorder[oi]))
                stack[-1].left = cur
                cur = None
                oi += 1
        return cur

如果二叉树中存在重复元素，上面的方法肯定行不通了，那怎么重建二叉树呢？

参考：http://blog.csdn.net/vividonly/article/details/6688327