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.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def minDepth(self, root: TreeNode) -> int:
if root is None:
return 0
if root.left is None and root.right is None:
return 1
l = self.minDepth(root.left)
r = self.minDepth(root.right)
# 如果左子树和右子树其中一个为空，那么需要返回比较大的那个子树的深度
if root.left is None or root.right is None:
return l + r + 1
# 左右子树都不为空，返回最小深度+1即可
return min(l, r) + 1/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if (root == null) return 0;
if (root.left == null && root.right == null) return 1;
int l = minDepth(root.left);
int r = minDepth(root.right);
if (root.left == null || root.right == null) return l + r + 1;
return Math.min(l, r) + 1;
}
}