111.minimum-depth-of-binary-tree

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111.minimum-depth-of-binary-tree

111. 二叉树的最小深度

给定一个二叉树，找出其最小深度。

最小深度是从根节点到最近叶子节点的最短路径上的节点数量。

说明: 叶子节点是指没有子节点的节点。

示例:

给定二叉树 [3,9,20,null,null,15,7],

3
   / \
  9  20
    /  \
   15   7

返回它的最小深度  2.

思路

参考 104

解法一

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def minDepth(self, root: TreeNode) -> int:
      if not root:
        return 0

      minleft = self.minDepth(root.left)
      minright = self.minDepth(root.right)

      if not root.left or not root.right:
        return minleft + minright + 1

      return min(minleft, minright) + 1

/**
 * 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 minDepth(root.right) + 1;
        if (root.right == null) return minDepth(root.left) + 1;
        return 1 + Math.min(minDepth(root.left), minDepth(root.right));
    }
}

/**
 * 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;
        int left = minDepth(root.left);
        int right = minDepth(root.right);
        return (left == 0 || right == 0) ? (left + right + 1) : Math.min(left, right) + 1;
    }
}