LeetCode解题之Minimum Depth of Binary Tree

原题

求一棵二叉树的最小高度，即从根节点到最近叶子节点的路径经过的节点数。

注意点：

• 无

例子:

输入:

    3
   /   9  20
    /     15   7
  /
 14

输出: 2

解题思路

可以通过树的广度优先遍历 Binary Tree Level Order Traversal 来实现，在广度优先遍历的过程中，每遍历一层就高度加一，如果某一个节点是叶子节点，那么当前的高度就是最小高度。

AC源码

# 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):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root is None:
            return 0
        depth, curr_level = 0, [root]
        while curr_level:
            depth += 1
            next_level = []
            for n in curr_level:
                left, right = n.left, n.right
                if left is None and right is None:
                    return depth
                if left:
                    next_level.append(left)
                if right:
                    next_level.append(right)
            curr_level = next_level
        return depth


if __name__ == "__main__":
    None

欢迎查看我的Github (https://github.com/gavinfish/LeetCode-Python) 来获得相关源码。