网站备案每年一次吗,网页个人主页模板,品牌软文营销案例,深圳做网站商1、首先写一个类表示二叉树
public class TreeNode {int num;TreeNode left;TreeNode right;public TreeNode(int num) {this.num num;}}2、根据前#xff0c;中序遍历#xff0c;在控制台我们可以得到两个结果pre 和 in#xff1a; /*** 前序遍历* param node*/public st…1、首先写一个类表示二叉树
public class TreeNode {int num;TreeNode left;TreeNode right;public TreeNode(int num) {this.num num;}}2、根据前中序遍历在控制台我们可以得到两个结果pre 和 in /*** 前序遍历* param node*/public static void PreTree(TreeNode node){if (node null){return;}System.out.println(node.val);PreTree(node.left);PreTree(node.right);}/*** 中序遍历* param node*/public static void MidTree(TreeNode node){if (node null){return;}PreTree(node.left);System.out.println(node.val);PreTree(node.right);}public static void main(String[] args) {TreeNode head new TreeNode(1);head.left new TreeNode(2);head.right new TreeNode(3);head.left.left new TreeNode(4);head.left.right new TreeNode(5);int[] pre {1, 2, 4, 5, 3};int[] in {2, 4, 5, 1, 3};}
3、接下来编写构建二叉树的方法
/*** 根据先序和中序遍历结果建出一颗树* 先序结果是pre[L1...R1], 中序结果是[L2...R2]*/public static TreeNode buildTreeNode(int[] pre, int[] in){if (pre null || in null || pre.length ! in.length){return null;}return f(pre, 0, pre.length-1, in, 0, in.length-1);}/*** int[] pre {1, 2, 4, 5, 3};* int[] in {2, 4, 5, 1, 3};*/public static TreeNode f(int[] pre, int L1, int R1, int[] in, int L2, int R2){if (L1 R1){return null;}// 根节点等于前序遍历的第一个数TreeNode head new TreeNode(pre[L1]);if (L1 R1){return head;}int find L2;while (in[find] ! pre[L1]){find;}head.left f(pre,L1 1, L1 find - L2, in, L2, find - 1);head.right f(pre, L1 find - L2 1, R1, in, find1, R2);return head;} 4、由于上面使用了while循环遍历该方法还可以进一步优化为
/*** 根据先序和中序遍历结果建出一颗树* 先序结果是pre[L1...R1], 中序结果是[L2...R2]*/public static TreeNode buildTreeNode1(int[] pre, int[] in){if (pre null || in null || pre.length ! in.length){return null;}HashMapInteger, Integer valueIndexMap new HashMap();for (int i 0; i in.length; i) {valueIndexMap.put(in[i], i);}return g(pre, 0, pre.length-1, in, 0, in.length-1, valueIndexMap);}public static TreeNode g(int[] pre, int L1, int R1, int[] in, int L2, int R2, HashMapInteger, Integer valueIndexMap){if (L1 R1){return null;}TreeNode head new TreeNode(pre[L1]);if (L1 R1){return head;}int find valueIndexMap.get(pre[L1]);head.left g(pre,L1 1, L1 find - L2, in, L2, find - 1, valueIndexMap);head.right g(pre, L1 find - L2 1, R1, in, find1, R2, valueIndexMap);return head;}
