剑指offer之重建二叉树

1 问题

重建二叉树：给定二叉树的先序遍历（根左右）和中序（左中右）遍历结果，建立这棵二叉树。输入保证二叉树无重复结点

以先序{1, 2, 4, 7, 3, 5, 6, 8}和中序{4, 7, 2, 1, 5, 3, 8, 6}为例

2 分析

先序遍历的特点，我们知道{1, 2, 4, 7, 3, 5, 6, 8}第一个元素1就是树的根节点，然后中序遍历{4, 7, 2, 1, 5, 3, 8, 6}的根节点在中间，因为我们从先序遍历里面得知1就是根节点，所以在中序遍历{4, 7, 2, 1, 5, 3, 8, 6}中，在中序遍历数组1的左边元素都是根节点左子树{4, 7, 2,}，这里是3个元素，所以在先序数组里面根节点1的后面3个元素也是左子树{2,4,7}，也是根节点的左子树，在中序遍历1的右边边元素都是{5, 3, 8, 6}都是根节点的右子树，然后我们在先序数组里面根节点1的后面第3个元素的后面到尾巴，也就是{3,5,6,8}也就是根节点的右子树。然后我们再把问题分解成构建左子树{2,4,7}和构建右子树{3,5,6,8}，以此递归处理。

我们构建左子树

再构建右子树

3 代码实现

import java.io.*;class Tree
{public int value;public Tree left;public Tree right;public Tree(int value){this.value = value;this.left = null;this.right = null;}/***前序遍历 */public void printTree(Tree node){if (null == node)return;System.out.println("value is " + node.value);printTree(node.left);printTree(node.right);}/**得到树的根节点*/public Tree getTree(int[] preDatas, int[] inDatas){if (null == preDatas || null == inDatas){System.out.println("preDatas is null or inDatas is null");return null;}Tree tree = buildTree(preDatas, 0, preDatas.length - 1, inDatas, 0, inDatas.length - 1);return tree;}/**构建树的左右子结构*/public Tree buildTree(int[] preDatas, int preStart, int preEnd, int[] inDatas, int inStart, int inEnd){if (null == preDatas || null == inDatas){System.out.println("preDatas is null or inDatas is null");return null;}System.out.println("preStart is:" + preStart + " preEnd is" + preEnd + " inStart is " + inStart + " inEnd is" + inEnd);//这里就是进行如果树的左子节点和右子节点是否为空进行设置if (preStart > preEnd || inStart > inEnd){return null;}Tree tree = new Tree(preDatas[preStart]);for (int i = inStart; i <= inEnd; ++i){   System.out.println("preDatas[preStart] is " + preDatas[preStart]);if (preDatas[preStart] == inDatas[i]){tree.left = buildTree(preDatas, preStart + 1, preStart + i - inStart, inDatas, inStart, i - 1);tree.right = buildTree(preDatas, preStart + i - inStart + 1, preEnd, inDatas, i + 1, inEnd);break;}}return tree;}
}class test  
{public static void main (String[] args) throws java.lang.Exception{int[] preDatas = {1, 2, 4, 7, 3, 5, 6, 8};int[] inDatas = {4, 7, 2, 1, 5, 3, 8, 6};Tree test =  new Tree(0);Tree tree = test.getTree(preDatas, inDatas);if (tree == null){System.out.println("tree is null");return;}//我们进行前序打印树        test.printTree(tree);}
}

4 运行结果

preStart is:0 preEnd is7 inStart is 0 inEnd is7
preDatas[preStart] is 1
preDatas[preStart] is 1
preDatas[preStart] is 1
preDatas[preStart] is 1
preStart is:1 preEnd is3 inStart is 0 inEnd is2
preDatas[preStart] is 2
preDatas[preStart] is 2
preDatas[preStart] is 2
preStart is:2 preEnd is3 inStart is 0 inEnd is1
preDatas[preStart] is 4
preStart is:3 preEnd is2 inStart is 0 inEnd is-1
preStart is:3 preEnd is3 inStart is 1 inEnd is1
preDatas[preStart] is 7
preStart is:4 preEnd is3 inStart is 1 inEnd is0
preStart is:4 preEnd is3 inStart is 2 inEnd is1
preStart is:4 preEnd is3 inStart is 3 inEnd is2
preStart is:4 preEnd is7 inStart is 4 inEnd is7
preDatas[preStart] is 3
preDatas[preStart] is 3
preStart is:5 preEnd is5 inStart is 4 inEnd is4
preDatas[preStart] is 5
preStart is:6 preEnd is5 inStart is 4 inEnd is3
preStart is:6 preEnd is5 inStart is 5 inEnd is4
preStart is:6 preEnd is7 inStart is 6 inEnd is7
preDatas[preStart] is 6
preDatas[preStart] is 6
preStart is:7 preEnd is7 inStart is 6 inEnd is6
preDatas[preStart] is 8
preStart is:8 preEnd is7 inStart is 6 inEnd is5
preStart is:8 preEnd is7 inStart is 7 inEnd is6
preStart is:8 preEnd is7 inStart is 8 inEnd is7
value is 1
value is 2
value is 4
value is 7
value is 3
value is 5
value is 6
value is 8