226. Invert Binary Tree

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本文介绍了一种翻转二叉树的算法实现，并通过具体的代码示例展示了如何使用递归方式来完成二叉树节点的左右子树互换过程。文章提供了完整的Python代码及运行结果。

226. Invert Binary Tree

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Discription:

Invert a binary tree.

     4  
   /   \  
  2     7  
 / \   / \  
1   3 6   9

         4  
       /   \  
      7     2  
     / \   / \  
    9   6 3   1

Analyze:

Code 1:

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


    #Generate a binary tree by a list ( like, [1,2,None,4,5,6,None,8,9,10,11,12] ) and return the root node.   
    def gen_Tree(val_list):
            if not val_list:
                return 
            root=TreeNode(val_list.pop(0))
            #print root.val
            le_list=[root]
            while val_list:
                tmp=le_list.pop(0)
                if val_list:
                    tmp.left=TreeNode(val_list.pop(0))
                    if tmp.left.val!=None:
                        le_list.append(tmp.left)
                if val_list:
                    tmp.right=TreeNode(val_list.pop(0))
                    if tmp.right.val!=None:
                        le_list.append(tmp.right)
                else:
                    return root
            return root


    # print the tree by level
    def show_Tree(root):
        def print_fun(le):
            for i in le:
                if i!=None:
                    print i.val,
                else:
                    print 'None',
            print ''
        cur_le=[root]
        next_le=[]
        print_fun(cur_le)
        while cur_le:        
            tmp=cur_le.pop(0)
            if tmp!=None:# and tmp.val!=None:
                if tmp.left:
                    next_le.append(tmp.left)
                else:
                    next_le.append(None)
                if tmp.right:
                    next_le.append(tmp.right)
                else:
                    next_le.append(None)
            if not cur_le:
                cur_le=next_le
                print_fun(cur_le)
                next_le=[]
                if not cur_le:
                    break            
        return

Solution code

    class Solution(object):
        def invertTree(self, root):
            """
            :type root: TreeNode
            :rtype: TreeNode
            """
            if root:
                root.right,root.left=root.left,root.right
                self.invertTree(root.left)
                self.invertTree(root.right)
            return root

    root=gen_Tree([1,2,None,4,5,6,None,8,9,10,11,12])
    show_Tree(root)

    1 
    2 None 
    4 5 None None 
    6 None 8 9 
    10 11 None None 12 None None None 
    None None None None None None

    s=Solution()
    root=s.invertTree(root)
    show_Tree(root)

    1 
    None 2 
    None None 5 4 
    9 8 None 6 
    None None None 12 None None 11 10 
    None None None None None None

Submission Result:

Status: Accepted
Runtime: 44 ms
Ranking: beats 79.29%