P18-二叉树最小深度-广度优先
作者:互联网
//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 * */ public class P18 { static class TreeNode{ int val; TreeNode left; TreeNode right; //记录当前深度 int deep; TreeNode(int val, TreeNode left, TreeNode right){ this.val = val; this.left = left; this.right = right; } } public static void main(String[] args) { TreeNode node7 = new TreeNode(7, null, null); TreeNode node6 = new TreeNode(6, node7, null); TreeNode node5 = new TreeNode(5, null, null); TreeNode node4 = new TreeNode(4, null, null); TreeNode node3 = new TreeNode(3, node6, null); TreeNode node2 = new TreeNode(2, node4, node5); TreeNode node1 = new TreeNode(1, node2, node3); System.out.println(minDepth(node1)); } //广度优先,从根节点向下计算,一层一层遍历,找到一个根节点就已经是最小深度 //问题是如果一层一层来?遍历2的时候,怎么做到再遍历3? //利用queue,先1进入queue,如果不是叶子,就把左右节点都入队, //然后左出队,不是叶子节点,又把孩子进队,右出队....... public static int minDepth(TreeNode root){ if(root == null){ return 0; } Queue<TreeNode> queue = new LinkedList<TreeNode>(); root.deep = 1; queue.offer(root); // while(!queue.isEmpty()){ TreeNode node = queue.poll(); if(node.left == null && node.right == null){ return node.deep; } if(node.left != null){ node.left.deep = node.deep + 1; queue.offer(node.left); } if(node.right != null){ node.right.deep = node.deep + 1; queue.offer(node.right); } } //不会走到这的,会再while中找到叶子节点就返回了 return 0; } }
标签:node,right,TreeNode,queue,二叉树,new,P18,广度,null 来源: https://www.cnblogs.com/YonchanLew/p/16129185.html