重建二叉树

【题目】根据二叉树的前序和中序遍历,重建二叉树,返回根节点。

1. 根据前序遍历,找到根节点。

2. 在中序遍历中找到根节点的位置,进而得到中序遍历中左右两个子树的划分位置。

3. 在前序遍历中找到左右两个子树的位置。

4. 根据这些位置,对左右子树分别进行递归操作,得到根节点的左右孩子节点。

  1 public class Main {
  2 
  3     public static void main(String[] args) throws Exception {
  4 
  5         int[] pre = { 1, 2, 4, 7, 3, 5, 6, 8 };
  6         int[] mid = { 4, 7, 2, 1, 5, 3, 8, 6 };
  7 
  8         Main main = new Main();
  9         Node root = main.process(pre, mid, 8);
 10 
 11         // 验证:前序遍历
 12         main.pre(root);
 13 
 14         System.out.println();
 15 
 16         // 验证:中序遍历
 17         main.mid(root);
 18 
 19         System.out.println();
 20 
 21         // 验证:后序遍历
 22         main.back(root);
 23     }
 24 
 25     public Node process(int[] pre, int[] mid, int length) throws Exception {
 26 
 27         if (null == pre || null == mid || length <= 0) {
 28             return null;
 29         }
 30 
 31         return buildTree(0, pre.length - 1, 0, mid.length - 1, pre, mid);
 32     }
 33 
 34     public Node buildTree(int startPre, int endPre, int startMid, int endMid,
 35             int[] pre, int[] mid) throws Exception {
 36 
 37         // 得到前序遍历第一个节点--根节点
 38         int rootValue = pre[startPre];
 39         Node rootNode = new Node(rootValue, null, null);
 40 
 41         if (startPre == endPre) {
 42             if (startMid == endMid && pre[startPre] == mid[startMid]) {
 43                 return rootNode;
 44             } else {
 45                 throw new Exception("invalid input");
 46             }
 47         }
 48 
 49         // 在中序遍历中找到根节点的位置
 50         int rootIndex = startMid;
 51         while (rootIndex <= endMid && mid[rootIndex] != rootValue) {
 52             rootIndex++;
 53         }
 54         if (rootIndex == endMid && mid[rootIndex] != rootValue) {
 55             throw new Exception("invalid input");
 56         }
 57 
 58         int leftLength = rootIndex - startMid;
 59         int preLeftEnd = startPre + leftLength;
 60 
 61         // 递归构建左子树
 62         if (leftLength > 0) {
 63             rootNode.setLeft(buildTree(startPre + 1, preLeftEnd, startMid,
 64                     rootIndex - 1, pre, mid));
 65         }
 66 
 67         // 递归构建右子树
 68         if (leftLength < endPre - startPre) {
 69             rootNode.setRight(buildTree(preLeftEnd + 1, endPre, rootIndex + 1,
 70                     endMid, pre, mid));
 71         }
 72 
 73         return rootNode;
 74     }
 75 
 76     public void pre(Node root) {
 77 
 78         if (null != root) {
 79             System.out.print(root.getValue() + " ");
 80         }
 81 
 82         if (null != root.getLeft()) {
 83             pre(root.getLeft());
 84         }
 85 
 86         if (null != root.getRight()) {
 87             pre(root.getRight());
 88         }
 89     }
 90 
 91     public void mid(Node root) {
 92 
 93         if (null != root.getLeft()) {
 94             mid(root.getLeft());
 95         }
 96 
 97         if (null != root) {
 98             System.out.print(root.getValue() + " ");
 99         }
100 
101         if (null != root.getRight()) {
102             mid(root.getRight());
103         }
104     }
105 
106     public void back(Node root) {
107 
108         if (null != root.getLeft()) {
109             back(root.getLeft());
110         }
111 
112         if (null != root.getRight()) {
113             back(root.getRight());
114         }
115 
116         if (null != root) {
117             System.out.print(root.getValue() + " ");
118         }
119     }
120 }
121 
122 class Node {
123     private int value;
124     private Node left;
125     private Node right;
126 
127     public Node(int value, Node left, Node right) {
128         this.value = value;
129         this.left = left;
130         this.right = right;
131     }
132 
133     public int getValue() {
134         return value;
135     }
136 
137     public void setLeft(Node left) {
138         this.left = left;
139     }
140 
141     public Node getLeft() {
142         return left;
143     }
144 
145     public void setRight(Node right) {
146         this.right = right;
147     }
148 
149     public Node getRight() {
150         return right;
151     }
152 
153 }
原文地址:https://www.cnblogs.com/jiangyi-uestc/p/7965834.html