Lintcode155-Minimum Depth of Binary Tree-Easy

155. 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.

Example

Example 1:

Input: {}
Output: 0

Example 2:

Input:  {1,#,2,3}
Output: 3	
Explanation:
	1
	  
	  2
	 /
	3    

 

注意：

如果用普通的递归，那么在helper方法之前要考虑只有一边树的情况（根结点+右子树 root.left == null && root.right != null /跟节点+左子树 root.right == null && root.left != null ）。一定要加&&，否则input为{2}(只有跟节点时)也会进入这个case。如果不考虑一边树的话，会误判左子树的长度为1，那么会minDepth会返回1.

递归法代码：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: The root of binary tree
     * @return: An integer
     */
    int minDepth = Integer.MAX_VALUE;
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        } else if (root.left == null && root.right != null) {
            helper(root.right, 2);
        } else if (root.right == null && root.left != null) {
            helper(root.left, 2);
        } else {
            helper(root, 1);
        }
        return minDepth;
    }
    public void helper(TreeNode root, int curDepth) {
        if (root == null) {
            return;
        }
        if (root.left == null && root.right == null) {
            if (curDepth < minDepth) {
                minDepth = curDepth;
            }
            return;
        }
        
        helper(root.left, curDepth + 1);
        helper(root.right, curDepth + 1);
    }
}

二分法：

1. 需要返回值

2. 自下而上求解（把问题分解成若干小问题）

public class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return getMin(root);
    }

    public int getMin(TreeNode root){
        if (root == null) {
            return Integer.MAX_VALUE;
        }

        if (root.left == null && root.right == null) {
            return 1;
        }

        return Math.min(getMin(root.left), getMin(root.right)) + 1;
    }
}