力扣111二叉树的最小深度（DFS）

Problem: 111. 二叉树的最小深度

题目描述

思路

1.欲望求出最短的路径，先可以记录一个变量minDepth，同时记录每次当前节点所在的层数currentDepth
2.在递的过程中，每次递一层，也即使当前又往下走了一层，则currentDepth++，当到达叶子节点时，比较并取出min【minDepth, currentDepth】
3.在归的过程中，因为是在往上层归，则currentDepth–；
4.返回最终的minDepth即可

复杂度

时间复杂度:

$O(n)$;其中$n$为二叉树的节点个数

空间复杂度:

$O(h)$；最坏空间复杂度

Code

DFS

/**
 * 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 {
     // record the minimum depth 
        private int minDepth = Integer.MAX_VALUE;
        // record the depth of the current node being traversed
        private int currentDepth = 0;

    public int minDepth(TreeNode root) {
       if (root == null) {
        return 0;
       }
       // start DFS traverssal from the root node
       travers(root);
       return minDepth;
    }

    private void travers(TreeNode root) {
        if (root == null) {
            return;
        }
    // increase the current depth when entering a node in the preorder position
    currentDepth++;

    // if the current node is a leaf, update the minimum depth
    if (root.left == null && root.right == null) {
        minDepth = Math.min(minDepth, currentDepth);
    }

    travers(root.left);
    travers(root.right);

    // decrease the current depth when leaving a node in the postorder position
    currentDepth--;

    }
}