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.
- Method 1:递归
- 这道题有一种特殊情况,当输入为[1, 2],我们返回2.
1 / 2
- 我们需要判断左右结点是否为null。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if(root == null) return 0;
if(root.left == null)
return Math.min(Integer.MAX_VALUE, minDepth(root.right)) + 1;
else if(root.right == null)
return Math.min(minDepth(root.left), Integer.MAX_VALUE) + 1;
else
return Math.min(minDepth(root.left), minDepth(root.right)) + 1;
}
}
还是在一刷的问题上卡了一下。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if(root == null) return 0;
if(root.left == null && root.right != null)
return Math.min(Integer.MAX_VALUE, minDepth(root.right)) + 1;
else if(root.left != null && root.right == null)
return Math.min(Integer.MAX_VALUE, minDepth(root.left)) + 1;
else
return Math.min(minDepth(root.left), minDepth(root.right)) + 1;
}
}
- Method 1: Recursion
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int minDepth(TreeNode root) { if(root == null) return 0; if(root.left == null && root.right == null) return 1; else if(root.left != null && root.right != null) return Math.min(minDepth(root.left), minDepth(root.right)) + 1; else if(root.left != null) return minDepth(root.left) + 1; else return minDepth(root.right) + 1; } }