[leetcode]#111. Minimum Depth of Binary Tree
2017-11-22 10:19
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输入一个二叉树,输出leaf节点的最小深度
解题思路:分几种情况考虑:1,树为空,则为0。 2,根节点如果只存在左子树或者只存在右子树,则返回值应为左子树或者右子树的(最小深度+1)。 3,如果根节点的左子树和右子树都存在,则返回值为(左右子树的最小深度的较小值+1)。
解题思路:分几种情况考虑:1,树为空,则为0。 2,根节点如果只存在左子树或者只存在右子树,则返回值应为左子树或者右子树的(最小深度+1)。 3,如果根节点的左子树和右子树都存在,则返回值为(左右子树的最小深度的较小值+1)。
# Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # @param root, a tree node # @return an integer def minDepth(self, root): if root == None: return 0 if root.left == None and root.right != None: return self.minDepth( root.right ) + 1 if root.left != None and root.right == None: return self.minDepth( root.left ) + 1 return min( self.minDepth( root.left ), self.minDepth( root.right ) ) + 1
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