leetcode 226. Invert Binary Tree

Invert a binary tree.

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

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

解法1:

本质是输的先序遍历

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution(object):def invertTree(self, root):""":type root: TreeNode:rtype: TreeNode"""# invert self.left# invert self.rightif not root: return Noneroot.left, root.right = root.right, root.leftself.invertTree(root.left)self.invertTree(root.right)return root

解法2，上述DFS的迭代解法：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution(object):def invertTree(self, root):""":type root: TreeNode:rtype: TreeNode"""        if not root: return Nonestack = [root]while stack:node = stack.pop()node.left, node.right = node.right, node.leftif node.right:stack.append(node.right)if node.left:stack.append(node.left)return root

解法3，使用BFS：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution(object):def invertTree(self, root):""":type root: TreeNode:rtype: TreeNode"""        if not root: return Nonenodes = [root]while nodes:nodes2 = []for node in nodes:node.left, node.right = node.right, node.leftif node.left: nodes2.append(node.left)if node.right: nodes2.append(node.right)nodes = nodes2return root

转载于:https://www.cnblogs.com/bonelee/p/8563696.html

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