[LeetCode] 111. Minimum Depth of Binary Tree Java
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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; } } } }
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