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  • 111. Minimum Depth of Binary Tree

    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.

    Note: A leaf is a node with no children.

    Example:

    Given binary tree [3,9,20,null,null,15,7],

        3
       / 
      9  20
        /  
       15   7

    return its minimum depth = 2.

     使用递归来做

    1.如果当前结点,为null。那么最小深度depth为0

    2.如果当前结点不为null,那么最小深度为1

       2.1 如果左结点和右结点都为null,那么到此结束,直接返回最小深度为1

       2.2 如果左结点和右结点都不为null,那么需要分别计算左结点和右结点的最小深度,取较小值+1

       2.3 如果左结点不为null,右结点为null,那么左结点的最小深度+1就作为,最小深度

       2.4 如果右结点不为null,左结点为null,那么右结点的最小深度+1就作为,最小深度

    public int MinDepth(TreeNode root)
            {
                if (root == null)
                {
                    return 0;
                }
    
                TreeNode left = root.left;
                TreeNode right = root.right;
                if (left == null && right == null)
                {
                    return 1;
                }
    
                int depth;
                if (left != null && right != null)
                {
                    int leftDepth = MinDepth(left);
                    int rightDepth = MinDepth(right);
                    depth = Math.Min(leftDepth, rightDepth);
                }
                else if (left != null)
                {
                    depth = MinDepth(left);
                }
                else
                {
                    depth = MinDepth(right);
                }
    
                return depth + 1;
            }

    简化版的

     public int MinDepth(TreeNode root)
            {
                if (root == null)
                {
                    return 0;
                }
    
                TreeNode left = root.left;
                TreeNode right = root.right;
    
                if (left == null)
                {
                    return MinDepth(right) + 1;
                }
    
                if (right == null)
                {
                    return MinDepth(left) + 1;
                }
    
                int leftDepth = MinDepth(left);
                int rightDepth = MinDepth(right);
                return Math.Min(leftDepth, rightDepth) + 1;
            }

    上面的解题思路,是深度优先。

    另外还有广度优先的算法

    https://blog.csdn.net/u011475210/article/details/79278219

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  • 原文地址:https://www.cnblogs.com/chucklu/p/10687035.html
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