2019-11-02 12:07:41 +08:00
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# 7. 重建二叉树
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[NowCoder](https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6?tpId=13&tqId=11157&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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根据二叉树的前序遍历和中序遍历的结果，重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
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2019-11-02 17:33:10 +08:00
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/31d9adce-2af8-4754-8386-0aabb4e500b0.png" width="300"/> </div><br>
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2019-11-02 12:07:41 +08:00
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## 解题思路
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前序遍历的第一个值为根节点的值，使用这个值将中序遍历结果分成两部分，左部分为树的左子树中序遍历结果，右部分为树的右子树中序遍历的结果。
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2019-11-02 17:33:10 +08:00
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/c269e362-1128-4212-9cf3-d4c12b363b2f.gif" width="330px"> </div><br>
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2019-11-02 12:07:41 +08:00
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```java
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// 缓存中序遍历数组每个值对应的索引
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private Map<Integer, Integer> indexForInOrders = new HashMap<>();
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public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
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for (int i = 0; i < in.length; i++)
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indexForInOrders.put(in[i], i);
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return reConstructBinaryTree(pre, 0, pre.length - 1, 0);
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}
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private TreeNode reConstructBinaryTree(int[] pre, int preL, int preR, int inL) {
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if (preL > preR)
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return null;
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TreeNode root = new TreeNode(pre[preL]);
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int inIndex = indexForInOrders.get(root.val);
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int leftTreeSize = inIndex - inL;
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root.left = reConstructBinaryTree(pre, preL + 1, preL + leftTreeSize, inL);
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root.right = reConstructBinaryTree(pre, preL + leftTreeSize + 1, preR, inL + leftTreeSize + 1);
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return root;
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}
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```
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2019-11-02 17:33:10 +08:00
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
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