重建二叉树

[](https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6?tpId=13&tqId=11157&tPage=1&rp=1&ru=%2Fta%2Fcoding-interviews&qru=%2Fta%2Fcoding-interviews%2Fquestion-ranking)

输入某二叉树的前序遍历和中序遍历的结果,请重建出该二叉树。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。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6},则重建二叉树并返回。

解题思路

  1. 通过前序遍历找到 root 节点
  2. 那么在 中序遍历中 root 节点的左侧则是左子树,右侧是右子树
  3. 依次类推,递归生成节点的左子树和右子树
  4. 构建过程由下往上

    public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
    Map<Integer, Integer> preIndex = new HashMap<>();
    for (int i = 0; i < pre.length; i++) {
        preIndex.put(pre[i], i);
    }
    
    return buildTree(preIndex, in, 0, in.length - 1);
    }
    
    private TreeNode buildTree(Map<Integer, Integer> preIndex, int[] in, int start, int end) {
    if (start == end) {
        return new TreeNode(in[start]);
    }
    int indexOfRoot = start;
    for (int i = start; i <= end; i++) {
        if (preIndex.get(in[i]) < preIndex.get(in[indexOfRoot])) {
            indexOfRoot = i;
        }
    }
    TreeNode root = new TreeNode(in[indexOfRoot]);
    if (start <= indexOfRoot - 1) root.left = buildTree(preIndex, in, start, indexOfRoot - 1);
    if (indexOfRoot + 1 <= end) root.right = buildTree(preIndex, in, indexOfRoot + 1, end);
    return root;
    }