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Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
题目链接:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
题目分析:根据先序找根,根据中序二分左右子树
5ms,时间击败64%
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildHelper(int[] preorder, int pos, int[] inorder, int left, int right) {
TreeNode root = new TreeNode(preorder[pos]);
if (left == right) {
return root;
}
for (int i = 0; i < inorder.length; i++) {
if (preorder[pos] == inorder[i]) {
int leftSize = i - left;
if (pos + 1 < inorder.length && left <= i - 1) {
root.left = buildHelper(preorder, pos + 1, inorder, left, i - 1);
}
if (pos + leftSize + 1 < inorder.length && i + 1 <= right) {
root.right = buildHelper(preorder, pos + leftSize + 1, inorder, i + 1, right);
}
break;
}
}
return root;
}
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder.length == 0) {
return null;
}
return buildHelper(preorder, 0, inorder, 0, inorder.length - 1);
}
}