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  • 889. Construct Binary Tree from Preorder and Postorder Traversal

    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[] and post[] are both permutations of 1, 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;
        }
    }
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  • 原文地址:https://www.cnblogs.com/fatttcat/p/10204159.html
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