一些二叉树相关算法Python的解决_二叉树算法python-优快云博客

本文链接：https://blog.youkuaiyun.com/progression/article/details/50487033

本文介绍了一种使用广度优先搜索算法来确定二叉树最小深度的方法，并通过示例代码展示了如何实现这一功能。此外，还探讨了二叉树的锯齿形层序遍历，包括广度优先及深度优先两种实现方式。

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二叉树的最小深度

广度优先

# 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 minDepth(self, root):
       if not root:
            return 0

        queue, level = [root], 1
        while queue:
            for i in xrange(len(queue)):
                node = queue.pop(0)

                if not node.left and not node.right:
                    return level

                if node.left:
                    queue.append(node.left)

                if node.right:
                    queue.append(node.right)

            level += 1

二叉树遍历用列表输入每层节点

广度有优先

# 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 zigzagLevelOrder(self, root):
        res, queue = [], [(root, 0)]
        while queue:
            curr, level = queue.pop(0)
            if curr:
                if len(res) < level+1:
                    res.append([])
                if level % 2 == 0:
                    res[level].append(curr.val)
                else:
                    res[level].insert(0, curr.val)
                queue.append((curr.left, level+1))
                queue.append((curr.right, level+1))
        return res

深度优先


def zigzagLevelOrder(self, root):
    ans = []
    self.dfs(root, 0, ans)

    for i in range(len(ans)):
        if i%2:
            ans[i] = ans[i][::-1] # reverse

    return ans

def dfs(self, root, level, ans):
    if not root:
        return

    if len(ans) < level+1: #should >= level+1
        ans.append([])

    ans[level].append(root.val)
    self.dfs(root.left, level+1, ans)
    self.dfs(root.right, level+1, ans)