/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
Stack<TreeNode> S = new Stack<TreeNode>(); List<List<TreeNode>> list = new List<List<TreeNode>>(); private void postNode(TreeNode node)
{
if (node != null)
{
S.Push(node);
if (node.left != null)
{
postNode(node.left);
}
if (node.right != null)
{
postNode(node.right);
} if (node.left == null && node.right == null)
{
list.Add(S.ToList());
}
S.Pop();
}
} public int MinDepth(TreeNode root)
{
postNode(root); var min = int.MaxValue; if (list.Count == )
{
min = ;
} foreach (var l in list)
{
var count = l.Count;
if (count < min)
{
min = count;
}
} return min;
}
}
https://leetcode.com/problems/minimum-depth-of-binary-tree/#/description
补充一个python的实现：
class Solution:
def minDepth(self, root: TreeNode) -> int:
if root == None:
return
left = self.minDepth(root.left)
right = self.minDepth(root.right)
if left == or right == :
return left + right +
return min(left,right) +