[LeetCode] 111. Minimum Depth of Binary Tree Java

题目：

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.

题意及分析：求一棵二叉树的最低高度，即根节点到叶子节点的最短路径。遍历二叉树，然后用一个变量保存遍历到当前节点的最小路径即可，然后每次遇到叶子节点，就判断当前叶节点高度和先前最小路径，取较小值作为当前最小路径。

代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int minDepth(TreeNode root) {
        if(root==null) return 0;
        int[] minHeight = {0};
        getHeight(minHeight,1,root);
        return minHeight[0];
    }

    public void getHeight(int[] minHeight,int height,TreeNode node){      //遍历树，得到每个点的高度
        if(node.left!=null){
            getHeight(minHeight,height+1,node.left);
        }
        if(node.right!=null){
            getHeight(minHeight,height+1,node.right);
        }
        if(node.left==null&&node.right==null){      //对根节点，做判断，看是否是当前最小
            if(minHeight[0]!=0){       //当前子节点高度和先前最小值相比
                minHeight[0]=Math.min(height,minHeight[0]);
            }else{      //第一次初始化
                minHeight[0]=height;
            }
        }
    }
}