889. Construct Binary Tree from Preorder and Postorder Traversal - Medium
阿新 • • 發佈:2019-01-01
Return any binary tree that matches the given preorder and postorder traversals.
Values in the traversals pre
and post
are distinct positive integers.
Example 1:
Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]
Note:
1 <= pre.length == post.length <= 30
pre[]
andpost[]
are both permutations of1, 2, ..., pre.length
.- It is guaranteed an answer exists. If there exists multiple answers, you can return any of them.
time: O(nlogn) ~ O(n^2), space: O(height)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * }*/ class Solution { public TreeNode constructFromPrePost(int[] pre, int[] post) { return buildTree(pre, 0, pre.length - 1, post, 0, post.length - 1); } public TreeNode buildTree(int[] pre, int pre_start, int pre_end, int[] post, int post_start, int post_end) { if(pre_start > pre_end || post_start > post_end) {return null; } TreeNode root = new TreeNode(pre[pre_start]); if(pre_start + 1 > pre_end) { return root; } int idx = 0; for(int i = post_start; i <= post_end; i++) { if(post[i] == pre[pre_start + 1]) { idx = i; break; } } root.left = buildTree(pre, pre_start + 1, pre_start + idx - post_start + 1, post, post_start, idx); root.right = buildTree(pre, pre_start + idx - post_start + 2, pre_end, post, idx + 1, post_end - 1); return root; } }