leetcode笔记--Minimum Depth of Binary Tree

Minimum Depth of Binary Tree
题目：难度（Easy）
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
Tags：Tree， Depth-first Search ，Breadth-first Search

法1：DFS

分析：此题不能直接套用Maximum Depth of Binary Tree的方法直接使用depthMin = min(leftMinDepth, rightMinDepth)+1。因为像单支树这样的树，按题意它的深度应该是子树的深度+1，而不是0。

代码实现：

<span style="font-size:14px;"># Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    #DFS    def minDepth(self, root):        """        :type root: TreeNode        :rtype: int        """        if root is None:            return 0        if root.left and root.right:            return min(self.minDepth(root.left), self.minDepth(root.right))+1        elif root.left:            return self.minDepth(root.left)+1        else:            return self.minDepth(root.right)+1</span>

法2：BFS

分析：找到的第一个叶子节点所在的高度就是该树的最小深度。

代码实现：

<span style="font-size:14px;"># 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 minDepth(self, root):        """        :type root: TreeNode        :rtype: int        """        if root is None:            return 0                    from collections import deque        queue = deque()        queue.append(root)        curNodesNumEachLevel = 1        depthMin = 1        while queue:            frontElem = queue.popleft()            curNodesNumEachLevel -= 1            #注意：判断是否是叶子节点的条件语句必须放在depthMin += 1的前面            #只有在某一层的全部节点考察完后深度才会加1，而在curNodesNumEachLevel=0之前，就可能先出现了叶子节点（在该层上）            if frontElem.left is None and frontElem.right is None:#找到叶子节点                break            if frontElem.left:                queue.append(frontElem.left)            if frontElem.right:                queue.append(frontElem.right)            if curNodesNumEachLevel == 0:                #上一层的节点都出队了，深度增加，到这一层                depthMin += 1                curNodesNumEachLevel = len(queue)        return depthMin</span>