/**
 * 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:

1 class Solution:
2     def minDepth(self, root: TreeNode) -> int:
3         if root == None:
4             return 0
5         left = self.minDepth(root.left)
6         right = self.minDepth(root.right)
7         if left == 0 or right == 0:
8             return left + right + 1
9         return min(left,right) + 1

 

Java版本:

 1 class Solution {
 2     
 3     public int minDepth(TreeNode root) {
 4         if(root == null){
 5             return 0;
 6         }
 7         if(root.left == null && root.right == null){
 8             return 1;
 9         }
10         int minDep = Integer.MAX_VALUE;
11         
12         if(root.left != null){
13             minDep = Math.min(minDepth(root.left),minDep);
14         }
15         
16         if(root.right != null){
17             minDep = Math.min(minDepth(root.right),minDep);
18         }
19         return minDep + 1;        
20     }
21 }

 

posted on 2017-04-24 21:34  Sempron2800+  阅读(121)  评论(0编辑  收藏  举报