leetcode111

/**
 * 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 == 0)
            {
                min = 0;
            }

            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 0
        left = self.minDepth(root.left)
        right = self.minDepth(root.right)
        if left == 0 or right == 0:
            return left + right + 1
        return min(left,right) + 1