输入某二叉树的前序遍历和中序遍历的结果,请构建该二叉树并返回其根节点。
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。
示例 1:
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
示例 2:
Input: preorder = [-1], inorder = [-1]
Output: [-1]
限制:
0 <= 节点个数 <= 5000
class Solution {/**map存放中序遍历对应的下标,然后根据前序遍历创建根节点,再递归创建左节点和右节点*/int[] preorder;Map<Integer, Integer> map = new HashMap<>();public TreeNode buildTree(int[] preorder, int[] inorder) {this.preorder = preorder;int n = preorder.length;for (int i = 0; i < n; ++i) map.put(inorder[i], i);return recur(0, 0, n - 1);}private TreeNode recur(int root, int left, int right) {if (left > right) return null;TreeNode node = new TreeNode(preorder[root]);int u = map.get(preorder[root]); //u是根节点对应的中序遍历下标//left为前序遍历 根节点下一个树,左右区间为中序遍历根节点对应下标前面node.left = recur(root + 1, left, u - 1);//u - left + 1为左子树的长度node.right = recur(root + u - left + 1, u + 1, right);return node;}}
